Physics 202 Test 2

(8c23p51) In the figure a sphere, of radius a = 14.2 cm and charge q = 6.00×10-6 C uniformly distributed throughout its volume, is concentric with a spherical conducting shell of inner radius b = 45.4 cm and outer radius c = 47.4 cm . This shell has a net charge of -q. Find expressions for the electric field, as a function of the radius r, within the sphere and the shell (r < a). Evaluate for r = 7.1 cm. Find expressions for the electric field, as a function of the radius r, between the sphere and the shell (a < r < b). Evaluate for r=29.8 cm. Find expressions for the electric field, as a function of the radius r, inside the shell (b < r < c). Evaluate for r = 46.4 cm. Find expressions for the electric field, as a function of the radius r, outside the shell (r > c). Evaluate for r = 48.4 cm. What is the charge on the outer surface of the shell?

(a) Construct a gaussian sphere inside the insulating sphere with radiuu r According to gauss law Electric field E = kqenclosed / r2 Here radius r = 0.02 m qenclosed = ( Total charge /total volume) * Volume of the gaussian sphere qenclosed = Qr3 / a3 = 7.5e-7 Then E = kQr / a3 ................(1) Here r = 0.071m , a = 0.142 m , Q = 6.00×10-6 C , k = 9*109 N.m2 /C2 plug all values in (1) we get E( r = 0.02 m) = 1.0125*1013 N/C ----------------------------------------------- E ( r = 0.06m ) = kQ / r2 Q = 3.6 C , k = 9*109 N.m2 /C2 E ( r = 0.06m) = 9.0*1012 N/C --------------------------------------------- The outer sphere is conductor the charge inside the conductor is zero hence electric field E( r = 0.09 m) = 0 .0 N/C ---------------------------------------------------------- (b) An equal and opposite charge induced on the inner surface of the outer conducting sphere The charge on the inner surface of a outer sphere is = - 3.6 C The charge on the outer surface of the outer conducting sphere is -2.0*10-6 + 3.60 q' = -3.599998 C The electric field at a distance r = 0.12 m is E = kq' / r2 ..............(3) k = 9*109 N.m2 /C2 plug all values in (3) we get E ( r = 0.12 m) = 2.249**1012 N/C please verify the unit of charge Q posted by u

A particle has a charge of +1.23 μC and moves from point A to point B, a distance of 0.200 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +8.70E-4 J. Calculate the magnitude of the electric force that acts on the particle.

(a) The work done on the electric charge is 8.70*10^-4; this is equal to the product of the magnitude of the constant force acting on it and the distance moved in the direction of the force. Hence the electric force is 8.70*10^-4/0.200 =4.35*10^-3 N in the direction from A to B.

For a charged hollow metal sphere with total charge Q and radius R centered on the origin: A. the charge is on the inside surface. B. only positive charges can be on the outside surface. C. the field inside the shell is zero. D. the field for r > R will be the same as the field of a point charge, Q, at the origin. E. inside the metal the field is strongest. F. the field on the outside is perpendicular to the surface.

(a) the charge is on the inside surface. False (b) only positive charges can be on the outside surface. False (C)the field inside the shell is zero. True. (d)the field for r > R will be the same as the field of a point charge, Q, at the origin. True (e) inside the metal the field is strongest. False (f)the field on the outside is perpendicular to the surface. True q(enclosed) = 0 for the Gaussian surface just inside the object only positive charges can be on the outside surface. False

(8c23p69) A thin, metallic, spherical shell of radius a = 6.0 cm has a charge qa = 9.00×10-6 C. Concentric with it is another thin, metallic, spherical shell of radius b = 10.80 cm and charge qb = 3.00×10-6 C. Find the electric field at radial points r where r = 0.0 cm. Find the electric field at radial points r where r = 8.4 cm. Find the electric field at radial points r where r = 16.2 cm. Discuss the criterion one would use to determine how the charges are distributed on the inner and outer surface of the shells. What is the charge on the outer surface of the outer shell?

1) at r = 0 cm = 0.0 m E = k*Qin/r^2 = k*qa/r^2 = 9*10^9*9*10^-6/0.0^2 = 2.25*10^7 N/c 2) at r = 8.4 cm = 0.084 m E = k*Qin/r^2 = k*qa/r^2 = 9*10^9*9*10^-6/0.084^2 =1.15×107 N/C 3) at r = 16.2 cm = 0.162 m E = k*Qin/r^2 = k*(qa + qb)/r^2 = 9*10^9*(1*10^-6 + 3*10^-6)/0.162^2 = 4.11×106 N/C 3) In elctrostatic equilibrium, -9*10^-6 charge is induced on inner surface of the outer shell to make elctric field inside the outer metal shell. so, charge on outer surface of the outer shell = 3*10^-6 + 9*10^-6 = 1.2*10^-5 C (make sure you use the calculator)

(8c25p27) In the figure, battery B supplies 6 V. Find the charge on each capacitor first when only switch S1 is closed. Take C1 = 1.3μ F, C2 = 2.7μ F, C3 =3.4μ F, and C4 = 4.8μ F. Q1? Q2? Q3? Q4? Find the charge on each capacitor later when switch S2 is also closed. Q1? Q2? Q3? Q4?

1. C13= (1.3e-6)(3.4e-6) /((1.3e-6)+(3.4e-6)) = 0.00000094 C24=((2.7e-6)(4.8e-6))/((2.7e-6)+(4.8e-6))=0.000001728 Q1=Q3=C13*V = 0.00000094 * 6 = 0.00000564 Q2=Q4=0.000001728 * 6 = 0.000010368 2. C12 = C1+C2= (1.3e-6)+(2.7e-6) = 0.000004 C34 = C3+C4= (3.4e-6)+(4.8e-6) = 0.0000082 V12=Q/C2=[ (C12*C34/C12+C34)V ]/C12 = 4.032 V34 = V - V12 = 6V - 4.032V = 1.968V C1=C1*V12=(1.3e-6)(4.032)= 0.0000052416 C2=C2*V12=(2.7e-6)(4.032)=0.0000108864 C3=C3*V34=(3.4e-6)(1.968)=0.0000066912 C4=C4*V34= (4.8e-6)(1.968)= 0.0000094464

The plates of parallel plate capacitor A consist of two metal discs of identical radius, R1 = 2.81 cm, separated by a distance d = 3.37 mm, as shown in the figure. a) Calculate the capacitance of this parallel plate capacitor with the space between the plates filled with air. b) A dielectric in the shape of a thick-walled cylinder of outer radius R1 = 2.81 cm, inner radius R2 = 1.31 cm, thickness d = 3.37 mm, and dielectric constant κ = 1.85 is placed between the plates, coaxial with the plates, as shown in the figure. Calculate the capacitance of capacitor B, with this dielectric. c) The dielectric cylinder is removed, and instead a solid disc of radius R1 made of the same dielectric is placed between the plates to form capacitor C, as shown in the figure. What is the new capacitance?

1. Ca = E0A/d simplyfied is Ca = E0(pieR^2)/d = (8.85e-12)(pie*2.81e-2^2)/3.37e-3 = 6.51e-11 2.Cb = [(Eo*pi)/d] * (K(R1^2 - R2^2)+R2^2 =[((8.85e-12)(pi))/(3.37e-3)]*(1.85((2.81e-2)^2-(1.3e-2)^2)+(1.3e-2)^2 =1.087e-11 3.Cc = [2E0(piR1^2)/d](K/K+1) =[(2(8.85e-12)(pi*(2.81e-2)^2)/3.37e-3](1.85/(1.85+1)) =8.457e-12 F

Find the charge on each capacitor later when switch S2 is also closed. Q1? Q2? Q3? Q4?

2. C12 = C1+C2= (1.3e-6)+(2.7e-6) = 0.000004 C34 = C3+C4= (3.4e-6)+(4.8e-6) = 0.0000082 V12=Q/C2=[ (C12*C34/C12+C34)V ]/C12 = 4.032 V34 = V - V12 = 6V - 4.032V = 1.968V C1=C1*V12=(1.3e-6)(4.032)= 0.0000052416 C2=C2*V12=(2.7e-6)(4.032)=0.0000108864 C3=C3*V34=(3.4e-6)(1.968)=0.0000066912 C4=C4*V34= (4.8e-6)(1.968)= 0.0000094464

(8c25p74) A slab of copper of thickness b = 1.194 mm is thrust into a parallel-plate capacitor of C = 9.00×10-11 F of gap d = 8.0 mm, as shown in the figure; it is centered exactly halfway between the plates. What is the capacitance after the slab is introduced?

C = ε A/d C' = ε A/(d-b) C' = C * (d/(d-b)) = 9*10^-11 * (8*10^-3/(8*10^-3-1.194*10^-3)) = 1.058e-10 F

the capacitance =

E0A/d

(8c25p39) A charged isolated metal sphere of diameter 24.0 cm has a potential of 8000 V relative to V = 0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.

E=V/r=8000/0.12=66666.66N/C Energy density=0.5 ε0 E2 = 0.0197 J/m3

electric flux (at an angle) = 'phi' =

EAcos(theta)

(c25p59_6e) Suppose that the negative charge in a copper one-cent coin were removed to a very large distance from Earth-perhaps to a distant galaxy-and that the positive charge were distributed uniformly over Earth's surface. By how much would the electric potential at the surface change? (See Sample Problem 22-7 in the problem supplement. By taking the mass of a penny to be 3.11 g, the positive or negative charge is found to be 1.37×105 C)

Electric potential due to point charge is V=kq/r where 'q' is the charge in coloumbs and 'k' is coloumb constant we know that radius of earth is r =6400 km by substitute above values we get V=(9*10^9)(-1.37*10^5)/(6400*10^3) =1.93*10^8 V

Three capacitors with capacitances C1 = 8.7 μF, C2 = 2.1 μF, and C3 = 4.1 μF are connected in a circuit as shown in the figure, with an applied potential of V. After the charges on the capacitors have reached their equilibrium values, the charge Q2 on the second capacitor is found to be 62. μC. a) What is the charge, Q1, on capacitor C1? b) What is the charge, Q3, on capacitor C3? c) How much voltage, V, was applied across the capacitors?

Find C12 which = C1 + C2 = 8.7 + 2.1 = 10.8 Find V which = Q2/Cs = 62/2.1 = 29.5 a) Q1 = C1*V = 8.7 * 29.5 = 256.65 uC or 0.00025665C b) Q3 = C12*V = 10.8 * 29.5 = 318.6 or 0.0003186 C c) C = (C12*C3)/(C3 +C12) = (10.8* 4.1)/(4.1+ 10.8) =2.97 so...V=Q3/C = 318.6/2.97

(8c25p15) A 210 pF capacitor is charged to a potential difference of 60 V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the measured potential difference drops to 45 V, what is the capacitance of this second capacitor? (pF)

Given that The capacitor of charge (C1) = 210 pF The potential difference of (V0)=60 V Consider the initial charge on the capacitor (q)=C1V0 =(210PF)(60V) =12600pC Now the after the battery is disconnected and teh secondcapaciotr wire is parallel to the first , the cahrge on the firstcapcaciotr is q1 =C1V The potential difference across the first capacitor drops to(V) =45 V, Then q1 =(210pF)(45V) =9450pC Since the charge is conserved in this process , then thecharge on the second capacitor is q2 =q -q1 =12600pC -9450pC =12150pC Then the capcacitance of the second capacitor is C2 =q2/V =12150pC/45V =70PF

(8c23p35) A square metal plate of edge length 8.0 cm and negligible thickness has a total charge of 4.00×10-6 C. Estimate the magnitude E of the electric field just off the center of the plate (at, say, a distance of 0.50 mm) by assuming that the charge is spread uniformly over the two faces of the plate. Estimate E at a distance of 30 m (large relative to the plate size) by assuming that the plate is a point charge.

In first case , plate can be considered as infinitely large, surface charge density sigma= q/A = 4e-6/(0.08*0.08) = 0.000625 C/m^2 Electric field is given by = sigma/2eo = 0.000625/(2*8.85e-12) = 3.53 *10^ 7 N/C In second case, it is point charge, E = kq/r^2 = 9e9*4e-6/30^2 = 40 N/C

(8c23p13) A point charge q = 8.30×10-6 C is placed at one corner of a cube of edge a = 0.30 m. What is the flux through each of the cube faces? See the figure. (Hint: Use Gauss' law and symmetry arguments.) What is the flux through side A? What is the flux through side B?

Let it be = total flux Then flux through the cube is equal to /8 (8 - the number of cubes to surround the charge) Flux through side A = 0 (from symmetry) Flux through side B = (/8) / 3 =/24 That is: [q/0 ] / 24 =(8.30×10-6 ) /(24*8.85*10-12 ) = 3.91*10^4 N*m2 /C

A charged isolated conductor :

No electrostatic charges can exist inside a conductor. All charges reside on the conductor surface

(8c25p8) In the figure find the equivalent capacitance of the combination. Assume that C1 = 10.0μ F, C2 = 4.7μ F, and C3 = 2.9μ F.

The capacitance of the series combination of C1 and C2 is C1*C2/(C1 + C2) = 10.0* 4.7/(10.0 + 4.7) =3.197 µF. This now acts as a single capacitor in parallel with C3, in which the capacitances add, so the total capacitance is 3.197+ 2.9 = 6.097µF

DEFINITION OF ELECTRIC POTENTIAL

The electric potential at a given point is the electric potential energy of a small test charge divided by the charge itself: V = EPE/q

The electric field inside a conductor

The electrostatic electric field E inside a conductor is equal to zero

Gauss' law can be formulated as follows:

The flux of E through any closed surface"!o = net charge qenc enclosed by the surface In equation form: E0*phi = qenclosed

An isolated charged conductor with a cavity :

There is no charge on the cavity walls. All the excess charge q remains on the outer surface of the conductor

Consider two separate systems, each with four charges of magnitude q arranged in a square of length L as shown above. Points a and c are in the center of their squares while points b and d are half way between the lower two charges. Select True or False for the following statements. The electric potential at b is zero. The electric field at b is zero. The electric field at c is zero. The electric field at d is zero. The electric potential at d is zero. The electric field at a is zero. The electric potential at c is zero. The electric potential at a is zero. Using the diagram above, find the magnitude of the electric field at point d. DATA: q= 0.750 μC, L= 0.40 m.

True False False False False True True True

(8c23p43) A planar slab of thickness of 2.00 cm has a uniform volume charge density of 1.20×10-2 C/ m3. Find the magnitude of the electric field at all points in space both inside and outside the slab, in terms of x, the distance measured from the central plane of the slab. What is the field for x = 0.50 cm? What is the field for x = 4.00 cm?

When x= 0.50cm that is inside the slab E=1.20*10-2*0.005/8.854*10-12 = 6.78*10^6 N/C When x= 4cm that is outside E= 1.20*10-2*0.02/2*8.854*10-12 = 1.36*10^7 N/C

(8c23p21) An isolated conductor of arbitrary shape has a net charge of +1.30×10-5 C. Inside the conductor is a cavity within which is a point charge q = +5.85×10-6 C. What is the charge on the cavity wall? What is the charge on the outer surface of the conductor?

a) -5.85 10-6 C (make sure its negative, or the opposite) b)(+1.30x10^-6)+(5.85x10^-6) =1.89*10-5 C

Need to figure out the one above

apply energy conservation Ui +KEi = Uf + KEf Kq^2/r1 + 0 = Kq^2/r2 + 1/2mv^2 9*10^9*(1.6*10^-19)^2/(7*10^-3) = 1/2*1.6726219 ×10-27 * V^2 + 9*10^9*(1.6*10^-19)^2/(0.017) => V^2 = 236.13 => V = 15.36 m/s

(8c23p74) Charge is distributed uniformly throughout the volume of an infinitely long cylinder of radius R = 2.00×10-2 m. The charge density is 8.00×10-2 C/ m3. What is the electric field at r = 1.00×10-2 m? What is the electric field at r = 4.00×10-2 m?

charge density, P = 8*10-2 C/m3 Radius = 2*10-2 m Electric field at distance r < R is given by: E = (P*R) /2E0 Electric field at r = 1*10-2 m, E1 = (P*r) /2*E0 = (8*10^-2)*(1*10^-2) /(2*8.85*10-12) = 4.52*10^7 N/C Electric field at distance r > R is given by: E = (P*r2) /(2*E0*R) Electric field at r = 4*10-2 m, E2 = (P*r2) /(2*E0*R) = ((8*10-2)*(4*10^-2)^2) /(2*(8.85*10^-12*)(2*10^-2)) = 3.62*10^8 N/C

(8c24p99) In the quark model of fundamental particles, a proton is composed of three quarks: two "up" quarks, each having charge +2e/3, and one "down" quark, having charge -e/3. Suppose that the three quarks are equidistant from one another. Take the distance to be 1.46×10-15 m and calculate the potential energy of the subsystem of two "up" quarks. (MeV) Calculate the total electric potential energy of the three-particle system. (MeV)

charge of two up quarks is q = 2e / 3 separation between any two quarks is 1.46×10-15 m potentail energy of two up quarks: U = Kq^2/r (9e9)((2/3) 1.6e-19)^2/ 1.46×10-15 = 7.014e-14 then divide that by the q again 1.6e-19 to get 4.384e5 finally move decimal over to get .4384 MeV the total electric potentail energy is 0 MeV the total consists of all pair-wise items U=K[(2e/3)(2e/3)/r +(2e/3)(-e/3)/r +(2e/3)(-e/3)/r] U=K[(4e^2/9r)-(4e^2/9r)] U=0

Two protons at rest and separated by 7.00 mm are released simultaneously. What is the speed of either at the instant when the two are 17.00 mm apart?

conserving energy V^2 = Kq^2/m (1/r1-1/r2) q = 1.6x10^-19 C m= 1.6726219 × 10-27 calculating we get V = 3.405 m/sec

The density of electric field vectors passing through a given area A is called the -________

electric flux 'phi' =EAcos(theta)

(c26p80_6e) Three capacitors are connected in parallel. Each has plate area A = 5.00×10-2 m2 and plate spacing d = 2.70×10-3 m. What must be the spacing of a single capacitor of plate area A if its capacitance equals that of the parallel combination? What must be the spacing if the three capacitors are connected in series?

here, area of each capacitor , A = 5 * 10^-2 m^2 spacing , d = 2.70×10-3 m capacitance of each capacitor , C = A*epsilon0/d as the capacitor are in parallel Ceq = C1 + C2 + C3 Ceq = 3*A*epsilon/d ....(1) area of plates of Ceq is A and let the spacing of Ceq be d' Ceq = A*epsilon0/d' from equation(1) 3*A*epsilon/d = A*epsilon0/d' d' = d/3 d'=(2.70×10-3)/3 the spacing when three capacitor are connected in parallel is 0.5 * 10^-3 m as the capacitor are in series 1/Ceq = 1/C1 + 1/C2 + 1/C3 Ceq = A*epsilon/3d ....(1) area of plates of Ceq is A and let the spacing of Ceq be d' Ceq = A*epsilon0/d' from equation(1) A*epsilon/3d = A*epsilon0/d' d' = 3d d' = 4.5 * 10^-3 m the spacing when three capacitor are connected in series is 4.5 * 10^-3 m

A parallel plate capacitor with a plate area of 11.0 cm2 and air in the space between the plates, which are separated by 1.9 mm, is connected to a 12.0-V battery. If the plates are pulled back so that the separation increases to 5.3 mm, how much work is done?

initial capacitance ci = Ae0/d = 11*10^-4*8.85*10^-12/1.9*10^-3 = 5.124e-12 pF INITIAL POTENTIAL ENERGY = 1/2civ^2 = (1/2)(5.124e-12)(12)^2 = 3.69e-10 J final capacitance cf = 11*10^-4*8.85*10^-12/5.3*10^-3 = 1.84e-12 pF final energy = 1/2cfv^2 = 1.32e-10 J WORK DONE (W) = U1-U2 =( 3.69-1.32)*10^-10 = 2.37e-10 J

E field is always _____ to the surface of a charged conductor

perpendicular

The electric field just outside the surface of a conductor is _____ to the surface at equilibrium under electrostatic conditions.

perpendicular

(8c24p13) What is the charge on the surface of a conducting sphere of radius 0.15 m whose potential is 200 V (with V = 0 at infinity)? What is the charge density on the surface of a conducting sphere of radius 0.15 m whose potential is 200 V (with V = 0 at infinity)?

potential due to a conducting sphere at a pont on the surface is V = k*q/r = (9*10^9*q)/0.15 = 200 9*10^9*q = 200*0.15 q = 200*0.2/(9*10^9) = 3.33*10^-9 C charge density is rho = Q/V V = ((4)*pi*r^2)) = (4)*3.142*0.15^2 = 2.83*10^-1 m^2 rho = Q/V = (3.33*10^-9)/(2.83*10^-1 ) = 1.18*10^-8 C/m^2 is the charge density

Is the slab pulled in or must it be pushed in? - pulled or pushed?

pulled

At equilibrium under electrostatic conditions, any excess charge resides on the _______ of a conductor.

surface

(8c25p74) A slab of copper of thickness b = 1.194 mm is thrust into a parallel-plate capacitor of C = 9.00×10^-11 F of gap d = 8.0 mm, as shown in the figure; it is centered exactly halfway between the plates. If a charge q = 8.00×10^-6 C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? How much work is done on the slab as it is inserted?

the ratio is = to d/(d-b) = 1.175 or just find work first and use the before and after values to find ratio by dividing them. the work: find the stored energy before: Q^2/2C =(8.00×10^-6)^2/2(9.00×10^-11) = 0.355 then find the stored energy after: Q^2/2C' = (8.00×10^-6)^2/2(1.058e-10) =0.302 then subtract before from after = 0.302-0.355= -0.053J

(8c23p3) A cube with 1.40 m edges is oriented as shown in the figure in a region of uniform electric field. A. Find the electric flux through the right face if the electric field, in newtons per coulomb ( N/C), is given by 4.20i. (N*m(shift-6)2/C). B. . Find the electric flux through the right face if the electric field, in newtons per coulomb ( N/C), is given by -2.20j. C. Find the electric flux through the right face if the electric field, in newtons per coulomb ( N/C), is given by -2.20i + 4.20k. D. What is the total flux through the cube for each of these fields?

the right side has an A vector = 1.4^2 j a) so flux= E*A = 4.20i * 1.4^2 j = 0 since i and j are orthogonal b) flux = E*A = -2.2j * 1.4^2 j = -4.312 c) flux = E*A (-2.20i + 4.20k) * 1.4^2 j = 0 since j is orthogonal to i and k d) since E is constant flux in = flux out so net flux=0 for all three

(8c25p24) The figure shows a varable "air gap" capacitor of the type used in manually tuned radios. Alternate plates are connected together; one group is fixed in position and the other group is capable of rotation. Consider a pile of n = 4 plates of alternate polarity, each having an area A = 1.00×10-2 m2 and separated from adjacent plates by a distance d = 2.20×10-3 m. Compute the maximum capacitance.

there are 4 plates so there are 3 capacitors so Ceq = 3C = 3(E0*A/d) = 3[(8.85e-12)(1.00*10-2 )/(2.20*10^-3)]= 1.207e-10

(8c24p64) Consider two widely separated conducting spheres, 1 and 2, the second having four times the diameter of the first. The smaller sphere initially has a positive charge q = 3.00×10-6 C, and the larger one is initially uncharged. You now connect the spheres with a long thin wire. How are the final potentials V1 and V2 of the spheres related? Find the final charges q1 and q2. q1? q2? What is the ratio of the final surface charge density of sphere 1 to that of sphere 2?

when we connect the two spheres with a wire, the potential on both spheres will be same. let R1 and R2 are the radii of the spheres. so, R2 = 4*R1 q1 + q2 = q ---(Equation 1) V1 = V2 k*q1/R1 = k*q2/R2 q1/R1 = q2/(4*R1) q1 = q2/4 ----(Equation 2) from equation 1 (plug in equation 2) q2/4 + q2 = q (times 4 on both sides) 5*q2 = 4*q q2 = (4/5)*q = (4/5)*3*10^-6 = 2.4*10^-6 C <<<<<----------------Answer q1 = q2/4 = 2.4*10^-6/4 = 0.6*10^-7 C <<<<<----------------Answer charge density is given by a equation but all i did was take the larger sphere q and divide by the smaller sphere q and got 4

(8c24p17) In the figure set V = 0 at infinity and let the particles have charges q1 = +q = -3.50×10-5 C and q2 = -5q. They are seperated by d = 30 cm, with q1 located at the origin. Locate any points on the x axis (other that at infinity) at which the net potential due to the two particles in zero. What is the positive position? What is the negative position?

you are given: d =30 cm q1 =-3.50×10-5 q2 = 7 x q1 . Since q1 is smaller, both points we are looking for must becloser to q1 than to q2. This means the positive position must bebetween the two charges. We know that the point must be 5 times farther from q2 than from q1, so... . x + 5x = d 6x =d x = 5cm is the positive position . Now for the negative positive, we know that the distance fromq1 is x and the distance from q2 is x +d so . x + d = 5x 4 x = d x = 7.5 cm so thenegative position is -7.5 cm --------------------------------- charges q1 = q = -3.50×10-5C q2 = -7q = -5 (-3.50×10-5 C) Sepration of the two charges r = 30 cm = 0.30 m If the potential of q1 = - potential of q2 then net potential zero. For positive positioni.e., in between q1 and q2 Kq1/x = -Kq2/(r-x) q1/x = -q2/(r-x) q1 / q2 = -x/(r-x) q /-5q = -x /(0.30 -x) -1/5 = -x/(0.30 -x) 0.30 -x = 5x 5x +x = 0.30 6x = 0.30 x = 0.30 / 6 = 0.05625m = 5.625 cm ------------------------------------------------------------------------------ For negative position Kq1/x = -Kq2/(r+x) q1/x = -q2/(r+x) q1 / q2 = -x/(r+x) q /-5q = -x /(0.30 +x) 1/5 = x/(0.30 +x) 0.30 +x = 5x 4x = 0.30 4x = 0.30 x = 0.30 / 4 = 0.075 m = 7.5 cm i.e., x = -7.5 cm

At equilibrium under electrostatic conditions, the electric field is ______ at any point within a conducting material.

zero