College Physics 10th edition chapter 17 & 18

The electric potential (relative to infinity) due to a single point charge Q is +400 V at a point that is 0.90 m to the right of Q. The electric potential (relative to infinity) at a point that is 0.90 m to the left of Q is

+400V

The electric potential (relative to infinity) due to a single point charge Q is 500V at a point that is 0.700m} to the right of Q.

+500V

charge of electron

-1.6 × 10^-19 C

Mass of proton

1.67 × 10^-27 kg

K equals?

9 x 10^9 N.m^2/C^2

Mass of electron

9.11 x 10^-31 kg

A point charge 4.20 μC is held fixed at the origin. A second point charge 1.30 μC with mass of 2.80×10−4 kg is placed on the x axis, 0.220 mfrom the origin. A. What is the electric potential energy U of the pair of charges? (Take U to be zero when the charges have infinite separation.) B. The second point charge is released from rest. What is its speed when its distance from the origin is 0.400 m .

A) U = k q1 q2 / d B) U - U' (U' = k q1 q2 / x(distance) )

A disk shaped parallel-plate capacitor has capacitance C. In terms of C, what would the capacitance be if the radius of each plate were increased by a factor of 3?

C = (πer^2)/d C = (πe3r^2)/d = 9C

Three negative point charges lie along a line as shown in the figure. Find the magnitude of the electric field this combination of charges produces at point P, which lies 6 m from the -2uC charge measured perpendicular to the line connecting the three charges. *Find the direction of the electric field due to each charge at point P, -2C at O, -5C is above O and at A. -5 C is below O and at B

E = Epo + E resultant -2 at O: Epo = (k x qo)/((OP)^2) qo = 2 OP = 6 EPA = EPB PA^2 = OA^2 + OP^2 qA = k(q/rA^2) E resultant = 2EAcos(a/2) cos(a/2) = OP/PA = 6/10

An electric field of magnitude E is measured at a distance R from a point charge Q. If the charge is doubled to 2Q and the electric field is now measured at a distance of 2R from the charge, the new measured value of the field will be:

E/2

A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 4 x 104 N/C. What work is done by the electric force when the charge moves (a) 0.45 m to the right; (b) 0.67 m upward; (c) 2.6 m at an angle of 45o downward from the horizontal?

F = Eq S = displacement (distance) a) W = FSCos90 = 0 b) W = FSCos0 c) W = FSCos135

Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = -3 nC and is at x = 4 cm . Charge q1 is at x = 2 cm .What is q1 (magnitude and sign) if the net force on q3 is zero?

F1 = F2 Q1 = q2(r1^2/r2^2)

Force of repulsion or attraction

F= k(q1q2/r^2)

Two small spheres spaced 20.0 cm apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57 x 10-21 N?

Find value of q. Then apply value in Ne = q/e(charge)

If a proton and an electron are released when they are 2.50×10^-10m apart (typical atomic distances), find the initial acceleration of each of them.

First find F= abs(k(q1q2/r^2)) Than apply in ae = F/Me ap = F/Mp

Two point charges are placed on the x-axis as follows: Charge q1 = 4 nC is located at x = 0.2 m, and charge q2 = -5 nC at x = 0.4m . What are the magnitude and direction of the net force exerted by these two charges on a negative point charge q3 = -0.6 nC placed at the origin?

Fnet = F13 + (-F23) = kq1q3/r13^2 - kq2q3/r23^2

Two point charges are located on the y-axis as follows: one charge q1 = -1.5nC located at y= -0.6m , and a second charge q2 = 3.2nC at the origin (y=0). What is the magnitude of the total force exerted by these two charges on a third charge q3 = 5nC located at y3 = -0.4m ?

Fnet = F23 + F13 = (kq2q3/r23^2) + (kq1q3/((r - r23) ^2)

If two electrons are each 1.50X10^10m from a proton,find the magnitude and direction of the net electrical force they will exert on the proton. angle 65 degrees

For magnitude first find F= k(q1q2/r^2). Then apply F in √2F²+2F²Cosθ. For direction Tanθ^-1(F1Sina/F1 + F1Cosa)

What are the magnitude and direction of the force each charge exerts on the other if they're equal charges?

Magnitde = F= k(q1q2/r^2)

If the potential (relative to infinity) due to a point charge is V at a distance R from this charge, the distance at which the potential (relative to infinity) is 2V is

R/2

Two unequal point charges are separated as shown. The electric field due to this combination of charges can be zero. __Region 1__+__Region 2__-__Region 3__

Region 3

What is the potential on point p -10 uc p +10uc

The potential (relative to infinity) is zero

If the electrical energy of two point charges is U when they are a distance d apart, their potential energy when they are twice as far apart will be

U/2

two point charges with charge +q are initially separated by a distance d. If you double the charge on both charges, how far should the charges be seperated for the potential energy between them to remain the same?

U2 = k2q2q/d = 4d

A small object carrying a charge of -8.00 nC is acted upon by a downward force of 20.0 nN when placed at a certain point in an electric field. a) what are the magnitude and direction of the electric field at the point in question? b) what would be the magnitude and direction of force acting on a proton placed at this same point in the electric field?

a) E = F/q = absI(20nN)/(-8nC)I = -2.5 N/C Magnitude : -2.5N/C Direction: Upward b) F = Eq = (2.5)(1.6x10^-19) Direction is upwards

A uniform electric field exists in a region between two oppositely charged plane parallel plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 3.2 cm away, in a time 1.5 x 10-8 s. (a) Find the magnitude of this electric field (b) Find the speed of the electron when it strikes the second plate

a) E = F/q. Find F by F = me x a. first find a = 2S/t^2

Electric fields in the atom.(a) within the nucleus. what strength of electric field does a proton produce at the distance of another proton, about 5.0 * 10^-15 m away? (b) at the electron. what strength of electric field does this proton produce at the distance of the electrons, approximately 5.0 * 10^-10 m away?

a) E = k(q/r^2) b) E = k(q/r^2)

In a rectangular coordinate system a positive point charge q =6 nC is placed at the point x = +0.150 m, y =0, and an identical point charge is placed at x = -0.150 m, y = 0.Find the x and y components, the magnitude, and the direction of the electric field at the following points: a) the origin; b) x =0.300 m, y = 0; c) x = 0.150 m, y = -0.400 m; d) x = 0, y = 0.200m.

a) Eo = E1 - E2 = 0.15 - 0.15 = 0 b) E = k[(q1/(r1+0.3)^2) + (q2/(r2+0.3)^2)] c) E1 = k(q/r1^2) r1 = sqrt(3^2 + -4^2) E1,x = E1cos θ E 1,y = E1sinθ E2 = k(q1/r2^2) E = sqrt((E1,x)^2 + (E1y E2y)^2) θ = 360 - Tan-1(E1y+E2y/E1x) d) E1 = k(q1/r1^2) r1 = sqrt(-0.15^2 + 0.2^2) E1,x = E1cos θ E 1,y = E1sinθ E2,x = -E1,x E 2,y = E1,y E = E1,x + E1y + -E1,x + E1,y direction upwards

A uniform electric field has magnitude E and is directed in the negative x direction. The potential difference between point a (at x= 0.60m ) and point b (at x= 0.90m ) is 240V. a) Which point, a or b, is at a higher potential? b) Calculate the value of E. c) Calculate the work done on the point charge by the electric field.

a) Higher potential is at point b because the electric field moves from higher potential to lower potential b) E = (Vb - Va)/d c) W = q(Vb - Va)

As you walk across a synthetic-fiber rug on a cold, dry winter day you pick up an excess charge of -55 uC. a) how many excess electrons did you pick up?b) what is the charge on the rug as a result of you walking across it

a) Ne = Q(charge)/e(charge) e = -1.6 X10^-19 b) charge on the rug is +55 uC

a) How much charge does a battery have to supply to a 5.00 uF capacitor to create a potential difference of 1.50V across its plates? b) How much energy is stored in the capacitor? c) How much charge would the battery have to supply to store 1.0 J of energy in the capacitor? d) What would be the potential across the capacitor in that case?

a) Q = CV b) U = 1/2CV^2 c) Q = sqrt(2CV) = sqrt(2(5uC)(1)) d) V = Q/C

A point charge with charge q1 = 2.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.30 μC moves from the point ( 0.150 m , 0) to the point( 0.250 m , 0.250 m ). a) What is the change in potential energy of the pair of charges? b) How much work is done by the electric force on q2.

a) U = U2 -U1 U1 = Kq1q2/r(first point) U2 = kq1q2/sqrt(r^2 (second point)) b) W = kq1q2(1/a-1/b) a = sqrt(x1^2+y1^2) b= sqrt(x2^2+y2^2)

Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the charges and moves along the line connecting them. What is the electric potential energy of the electron when it is a) at the midpoint? b) 10.0 cm from the 3.00 nC charge?

a) U = k x qe((q1/r1^2) + (q2/(r2^2)) b) U = k x qe((q1/r1^2) + (q2/(r2^2)) r1 = 0.1m r2 = 0.4m

A 12.5μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. a.) How much energy is stored in the capacitor before the dielectric is inserted? b.) How much energy is stored in the capacitor after the dielectric is inserted? c.) By how much did the energy change during the insertion? Did it increase or decrease?

a) V0 = 1/2C0V0^2 b) V = 1/2CV^2 C = kC0 VΔ= V - V0

Two point charges q1 = 2.40nC and q2 = -6.50nC are 0.100 m apart. Point A is midway between them; point B is 0.080 m from q1 and 0.060 m from q2 (the figure (Figure 1) ). Take the electric potential to be zero at infinity. a) Find the potential at point A b) Find the potential at point B c) Find the work done by the electric field on a charge of 2.40nC that travels from point B to point A

a) VA = k[q1/r1 + q2/r2] b) VB = k[q1/r1B + q2/r2B] c) W = Vq V = VA - VB

A parallel-plate air capacitor has a capacitance of 500 pF and a charge of magnitude 0.200uC on each plate. The plates are 0.600mm apart. a) What is the potential difference between the plates? b) What is the area of each plate? c) What is the electric-field magnitude between the plates? d) What is the surface charge density on each plate?

a) Vab = Q/c b) A = Qd/(e0 x Vab) e0 = 8.854 x 10^-12 c) E = Vab/d d) Q/A

A parallel-plate capacitor having plates 6.00cm apart is connected across the terminals of a 12.0V battery. a) Being as quantitative as you can, describe the location of the equipotential surface that is at a potential of 6.00V relative to the potential of the negative plate. Avoid the edges of the plates. b) Do the same for the equipotential surface that is at 2.00V relative to the negative plate. c) What is the potential gradient between the plates?

a) a +6v equipotential surface will be at the middle of the capacitor plates. The distance between the plates are 6cm so the +6V will be at a distance of 3cm from either plates and will be parallel from the plates and perpendicular to the electric lines of force. b) the equipotential surface at +2V is parallel to the negative and 1cm from it c) Ex = -V/x = -12/.06

A potential difference of 4.75 kVis established between parallel plates in air. a) If the air becomes electrically conducting when the electric field exceeds 3 uC V/m, what is the minimum separation of the plates?

a) d = (Va - Vb)/E = 4.75 kV/3uC V/m

A negative charge of -0.55μC exerts an upward 0.2N force on an unknown charge 0.3m directly below it. a) what is the unknown charge? b) what is the magnitude and direction of the force that the unknown charge exerts on the -0.55μC charge?

a) q2 = Fr^2/Kq1 sign of charge is positive b) The direction is downward since the unknown charge attracts -0.55μC toward it.

acceleration of electron

ae = F/Me

acceleration of proton

ap = F/Mp

Three equal point charges are held in place as shown in the figure If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? +__d__+(q1)___2d___+(q2)

d+2d = 3d F1 = (3)^2F2 = 9F2

If the electric field is at E at a distance d from a point charge, it's magnitude will be 2E at a distance

d/sqrt(2)

How many excess electrons must you add to an object to give it a charge of -2.5 uC?

n=q/e e = -1.6 x 10^-19

quantization of charge

q=ne n is number of electrons. e is electon charge (-1.6x10^-19)

The repulsive force between two electrons has a magnitude of 16.0 N. What is the distance between the electrons?

r = √(kq1q2/F) q=charge of electron = -1.6 x 10^-19

A small particle has charge -5.00 and mass 2.0×10−4 kg . It moves from point A, where the electric potential is VA = +200 V , to point B, where the electric potential VB = +800 V is greater than the potential at point A. The electric force is the only force acting on the particle. The particle has a speed of 5.00 m/s at point A. What is its speed at point B? Is is moving faster or slower at B than at A?

v = sq rt [ u2 +2q (VBA) /m ] VBA = (VB - VA) = 200-800=-600 the particle moves faster at point B than Point A