Physics 2049

Exam 1

Spring 2002

r12

q1

q2

q3r13

1. In the figure q1 = 300µC, q2 = 400µC, q3 = 500µC, r12 = 9m, r13 =12m. Compute the magnitude of the total force on q3.

~F3NET = ~F31 + ~F32

=kq3q1

r213

i +kq3q2

r212 + r2

13

(

cos θi− sin θj)

with θ = arctan(r12

r13) = 36.87◦

~F3NET = 15.8Ni− 4.8Nj

|~F3NET | =√

(15.8)2 + (4.8)2N = 16.5N

1

q q

q

Q

2. Three identical charges q are placed at the corners of the equilateraltriangle of side length L. The fourth charge Q is placed in the centerof the triangle. Find the value of Q in terms of q that would put thewhole system of four charges in equilibrium.

Calculating the net force on the top q and setting it to zero to enforceequilibrium we get,

~FqNET = −k|qQ|x2

j +2kq2

L2sin(60◦)j = 0

where x is the distance between q and Q. In terms of L,

x =(L/2)

cos(30◦)= L/

√3 (1)

so that

3k|qQ|L2

=

√3kq2

L2

or Q = −q/√

3 (must have opposite sign for equilibrium)

e

E

2

3. An electron is placed in an electric field as shown. Find the directionof the electrostatic force on the electron.

F = qE = −eE

Since F and E are antiparallel the force is directed to the left, ←.

4. A particle of mass 1 mg, and charge 6µC traveling with constantvelocity ~v0 = 10m/si enters a region with a constant electric field~E = 1.2N/Cj. What is the speed 2 seconds after it enters the regionwith the electric field?

The velocity of the charge only changes in the j direction,

F = qE = may ⇒ ay = qE/m = 7.2m/s2

vy(t) = ayt⇒ vy(2s) = 14.4m/s

|~v(2s)| =√

(10)2 + (14.4)2)m/s = 17.53m/s

5. A spherical nonconducting balloon has a positive charge distributeduniformly over its surface. If the balloon is expanded (no extra chargeis added) the magnitude of the electric field inside the balloon , andthe magnitude of the electric field near the outer surface . If a pointP is outside the balloon before expansion, and inside after expansion,the magnitude of the E field at that point .

Use Gauss’s law to get the electric field,∮

S

~E · d~a = E(4πr2) =qin

ε0

E =kqin

r2

Since all the charge is uniformly distributed on the surface of the bal-loon, the electric field inside is zero. Therefore, the field after expansionwill remain zero. A point just outside the surface of the balloon theelectric field will decrease since the electric field drops off as 1/r2. Itshould also be evident that since the electric field is positive outside

the balloon, any point that starts outside and ends inside where thefield is zero will experience a decrease.

3

x=2cm x=12cm

x

z

y

6. A cube with sides of length 10 cm rests on the y = 0 plane, with oneedge along the line x = 2cm and another edge along x = 12cm, asshown in the above figure. A nonuniform field (in N/C) pierces the thecube and is described by:

E(x) = Axi

where A =7.0E+6 N/(C·m). How much charge (in C) is inside thecube?

Electric field flux only occurs at the x = 2cm and x = 12cm faces whichwe will call left and right, respectively. Gauss’s law tells us,

∮

S

~E · d~a =Q

ε0

=∫

right

~E · d~a +∫

left

~E · d~a

The differential area elements on these surfaces are d~aright = dydzi andd~aleft = −dydzi.

Q = ε0 ×(

E(x)

∣

∣

∣

∣

x=12cm×L2 − E(x)

∣

∣

∣

∣

x=2cm×L2

)

where L is the side length. Therefore,

Q = ε0 × 7000Nm2/C = 6.2E− 8C

4

7. The electric field at the surface of a sphere of radius R is measured.It is found to be radially outward and to have the same magnitude allover the sphere. Call that magnitude E1. A second sphere of radius2R is found to have a radially inward electric field that is uniform alsobut with magnitue 4E1. What is the ratio of the charge in the secondsphere to the charge in the first sphere (Q2/Q1)?

If we place Gaussian surfaces on the surfaces of both spheres they wouldtell us the magnitudes of the electric field are,

E1 =k|Q1|R2

1

, E2 =k|Q2|R2

2

−→ E1

E2=

R22|Q1|

R21|Q2|

=1

4=

4|Q1||Q2|

Since the fields point in opposite directions, the signs of Q1 and Q2 areopposite and we have

Q2

Q1= −16

8. A clever physicist produces a very long, cylindrical distribution of elec-trical charge. The diameter of the cyclindrical distribution is D. Insidethe volume charge density varies radially as,

ρ(r) = ρ0

(

1− 2rD

)

For r > D/2, ρ(r) = 0. Here r is the radial distance from the cylindricalaxis and ρ0 is a negative constant ρ0 < 0. Calculate the electric fieldgenerated by rho for all r > D/2. That is, at any point outside of theregion containing the charge.

Again we use Gauss’s law, but this time for cylindrical symmetry,∮

S

~E · d~a = E(2πrL) = qin/ε0

qin =∫

dV ρ(V ) =∫ 2π

0dφ∫ L

0dz∫ D/2

0dr rρ(r)

= 2πL|ρ0|∫ D/2

0dr r

(

1− 2r

D

)

5

= 2πL|ρ0|(

D2

8− D2

12

)

= πL|ρ0|D2

12

−→ ~E =|ρ0|D2

24ε0rinward

Now by the way of the 5 possible answers, only 2 had the electric fielddirected inward, allowing you to eliminate 3 of the possible choices. Ofthe remaining two, only the one derived above has the correct dimen-sions of of an electric field. Therefore, one could have answered thisquestion without having to even know what an integral is!

9. The electric field in megaN/C at a point halfway between a point chargeof +2µC at the origin and a point charge of -3µC at x = +10 cm is:

Between the two charges, the electric field is pointed to the right andthe components add,

E(x) =k|q1|x2

+k|q2|

(0.1m− x)2

E(0.05m) =k(2E-6C+ 3E-6C)

(0.05m)2= 1.8E7 N/C = 18 megaN/C

10. An electric dipole is composed of equal and opposite charges each ofmagnitude 0.1µC, placed at y = −1 cm, respectively. The magnitude ofthe electrostatic field due to this dipole (in N/C) at the point x = +3mis:

The magnitude of the electric field along the z axis due to the twocharges along the y axis comes from adding the y components andcancelling the z components to give,

E(z) =2kq

(d/2)2 + z2sin θ =

kqd

((d/2)2 + z2)3/2

where sin θ = d/2√(d/2)2+z2

, and d is the distance between the charges.

The dipole field comes from approximating the electric field for z � d,so we can neglect the d/2 in the denominator and we are left with,

E(z) ≈ kqd

z3= k

(1E-7C)(0.02m)

(3m)2= 0.67 N/C

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