Physics II Exam 1

Given that q=+12uC and d=19 cm (a) find the direction and magnitude of the net electrostatic force exerted on the point charge q2 (b) How would your answers to part a change if the distance d were tripled?

(a) F2=k/d^2 (-q1*q2++q1*q3)xhat 4.0 X (8.99x10^9)(12x10^-6)^2/(0.19m)^2= 140 N= 0.14kN toward q (b) the magnitude would be cut to a ninth

Two point charges, the first with a charge +3.13x10^-6 C and the second with a charge of -4.47x10^-6 C, are separated by 25.5 cm. (a) find the magnitude of the electrostatic force experience by the positive charge (b) Is the magnitude of the force experienced by the negative charge greater, less, or the same as that experienced by the positive charge?

(a) F=k*|q1||q2|/r^2 (8.99*10^9)* (3.13*10^-6)(4.47*10^-6)/ (0.255m)^2 = 1.93 N (b) same as that experience by the positive charge

Suppose the conducting shell in the figure which has a point charge +Q at its center has a nonzero net charge. How much cahrge is on the inner an douter surface of the shell when the net hcarge of the shell is (a) -2Q (b)-Q (c) +Q

(a) inner surface: -Q and Outer surface= -Q (b) inner surface= -Q and outer surface = 0 (c) inner surface=-Q and outer surface +2Q

The electric field lines surrounding three charges are shown in the figure. The center charge is q2=-10.0uC (a) What are the signs of q1 and q3 (b) find q1 (c)find q3

(a) positive (b) 5.00 uC (c) the magnitude of q3 = 5.00uC

a surface encloses the charges q1=3.2 uC, q2=6.9 uC, and q3= -4.1 uC. Find the electric flux through this surface

= (q1+q2+q3)/E0 = (3.2+6.9-4.1)x10^-6/ 8.85x10^-12 =6.8x10^5

two point charges lie on the x axis. A charge of +6.2uC is a the origin, and a charge of -9.5uC is a tx=10.0 cm. What is the net electric field at (a) x=-4.0 cm (b) x= +4.0

=( |q1|/r1^2 + |q2|/r2^2 ) x hat (8.99x10^9)[-6.2x10^-6/(0.040)^2 + 9.5x10^-6/ (0.140)^2 ] x hat = a=-3.0x10^7 N/C =( |q1|/r1^2 + |q2|/r2^2 ) x hat (8.99x10^9)[6.2x10^-6/(0.040)^2 + 9.5x10^-6/ (0.060)^2 ] x hat = a=-5.9x10^7 N/C

A uniform electric field with a magnitude of 1200 points in the neagtive x direction as shown in the figure (a) What is the difference in electric potential, change in V= Vb-Va between points A and B? (b) What is the difference in electric potenial change V =Vb-Vc between points B and C? (C) What points C and A? (d) from the information given in this problem, is it possible to determie the value of the electric potenial at A

A)change in V = Vb-Va= -E change in x = 0 B) change in V = -E(xb-xc) -(-1200)(-0.040)= -48 V C) Change in V= -E(xc-xa) -(-1200)(0.040)

Object with a charge of -3.6uC and a mass of 0.012 kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight (a) find the direction and magnitude of the electric field (b) if the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration

E= mg/q (0.0121 kg)(9.81)/ 3.6x10^-6 =3.3x10^4 E=E(-y)= -3.3x10^4 a=gy= 9.81 m/s^2 y hat

a +5.0 uC charge experiences a 0.44 N force in the positive y direction. If this charge is replaced with a -2.7 uC charge, what force will it experience?

E=F/q =0.44/5.0x10^-6= 8.8x10^4 F=qE= (-2.7x10^-6)(8.8x10^4)= -24 N

When adhesive tape is pulled from a dispenser, the detached tape acquires a positive charge and the remaining tape in the dispenser acquires a negative charge. If the tape pulled from the dispenser has 0.14uC of charge per centimeter, what length of tape must be pulled to transfer 1.8x10^13 electrons to the remaining tape?

L=N(e)/(q/l) ex) (1.8x10^13)(1.60x10^-19)/0.14x10^-4= 21 cm

A charge of 3.05uC is held fixed at the origin. A second charge of 3.05 uC is released from rest at the position (a) if the mass of the second charge is 2.16 g, what is its speed with it move infintely far from the origin (b) at what distance from the origin does the seocnd charge attain half the speed it whill have at infinity

Vf= square root (2kq^2/mr) square root ( 2(8.99x10^9)(3.05x10^-6)^2/ (0.00216 kg)sqr root (1.25)^2+(0.570)^2 = 7.51 m/s b) r=4/3 sqr root (1.25)^2+(0.570)^2=1.83 m

it is given that the electric potential si zero at the center of the square in the figure (a) if q1=+Q what is the value of the charge q2? (b) is the electric potenital at point A postivie negative, or zero? (c) is the electric potential at point B positive, negative, or zero

a) -Q b) positive c) negative

A charge of -2.205 uC is located at (3.055, 4.501m) and a charge of 1.800 uC is located at (-2.533m, 0) (a) findd the electric potential at the origin (b) three is one point on the line connecting these two hcarges where the potential is zero. Find this point

a) V= kq1/r1 +kq2/r2 (8.99-10^9)[-2.205*106-6/sqr root (3.055)^2 + (4.501)^2 + 1.800x10^-6/2.533] = 2.74 kV b) d= sqr root [3.055- (-2.533)]^2 +(4.501 m)^2 = 7.175 m x= q2d/ q2-q1 = (1.800*10^-6) (7.175)/1.800*10^-6 - (-2.205x10^-6)= 3.225 m x0= x/d= (x1-x2)+x2= -0.021 y0 = x/d= (y1-y2)+y2= 2.023 m (-0.021, 2.023)

Find the potential difference required to acclerate protons from rest to 10% of the speed of light

change in V = mv^2/ 2e (16.10^-27kg)(0.1)^2(3x10^8/2(1.6x10^-19)= 5MV

Referring to the figure suppose q2 is not known. Instead it is given that q1+q2 =-2.5uC. Find q1, q2, and q3

q1+q2 =-2.5uC (-1/2 * q2) +q2= -2.5 q2 = -5.0 q1 and q3= 1/2(-5.00)= 2.50

a system consists of two positive point charges, q1 and q2>q1. The total charge of the system is +62.0uC and each charge experiences an electrostatic force of magnitude 85.0 N when the separation between them is 0.270 m. Find q1 and q2

q1=q+- sqr root (Q^2-4(1)(d^2*F/k)/(2)(1) 62.0 X10^-6+- sqr root (62.0x10^-6)^2-4(0.270)^2(85/8.99x10^9)/2=31.0*10^-6 +- 16.5x10^-6 q2= 47.5 q1= 14.5 uC

A sphere of radius 4.22 cm and uniform surface charge density +12.1uC/m^-2 exerts an electrostatic force of magnitude 46.9x10^-3 N on a point charge of +1.95uC. Find the separation between the pint charge and the center of the sphere

r=sqr root (4*pi*k*q*o*R^2)/(F) sqr root (4pi(8.99x10^9)(1.95x10^-6)(12.1x10^-6)(0.0422)^2/46.9*10^-3 = 0.333 m = 33.3 cm

A typical 12 V car battery can deliver 7.5 x10^5 C of cahrge. If the energy supplied by the battery could be converted entirely to kinetic energy, what speed would it give to a 1400 kg car that is initally at rest

v= square root (2q*change in V)/ m square root (2(7.5x10^-5)(12)/1400 kg = 110 m/s = 0.11 km/s