Sheet 2 Electric Potential

1- Prove that the electric potential at a point on the axis of a hollow disc with internal radius (a) and external radius (b) is given as:

,

where is charge per unit area of the disc and x is the distance between the point and disc center. A hollow disc, with internal radius (5 cm) and external radius (8 cm), is uniformly charged with (Q = 49 nC). When a proton starts from rest at a point on the disc axis with distance (10 cm) from its center, find its velocity when it reaches a point on the disc axis with distance (15 cm) from its center.

2- In the shown figure, find the electric

potential at the point (P) and calculate also the total potential energy for the whole system of charges.

3- A uniformly charged rod with length (6 cm) have a charge (5 C). Find the electric

potential energy between the charged rod and a point charge (q = 2 C) which is placed at a point along the axis of the rod with distance (9 cm) from the near end of the rod.

4- Consider two thin, conducting, spherical shells as shown in Figure. The inner

shell has a radius (a = 25 cm) and a charge of (Q1 = 8.0 nC). The outer shell

has a radius (b = 30 cm) and a charge of (Q2 = 15.0 nC). Find (a) the electric

field E and (b) the electric potential V in regions A, B, and C, with V = 0 at

r = .

Q4 = 2 nC

Q1 = 4 nC Q2 = -6 nC

Q3 = -8 nC

P

10 cm

6 cm

cmc

mcm

b a

C

B

A

5- In the shown figure, find the potential at (A) and (B). When a proton is projected from point (B) with velocity (3×106 m/s) find its velocity when it reaches point (A). (mp = 1.67×10-27 kg and qp = 1.6×10-19 C). If the proton is projected from point (A) with velocity (3×106 m/s) find its velocity when it reaches point (B).

6- Two point charges (Q1 = 4 C, Q2 = -9 C) are placed so that the distance between them is (80 cm). Find a- the point at which the electric field intensity is zero and find the electric potential

at this point. b- the point at which the electric potential is zero and find the electric field intensity

at this point.

7- Two conducting spheres are separated from each other by a long distance. The first

has a charge (6 C) and radius (30 cm) and the second has a charge (-4 C) and radius (50 cm). The spheres are connected together by a conducting wire. Find: a- The charge per unit area for each sphere after connection. b- The quantity of charge transferred between the spheres after connection. c- The change in potential at the surface for each sphere.

8- A conducting plate has a large area ( plate) is charged with a charge per unit area

( = 5 nC/m2). Find the potential difference between points (A) and (B) as shown in figure. If an proton starts motion from rest at point (A) find its velocity when it reaches point (B).

50 cm

A B

Q1 = 1.5 C Q2 = - 1.5 C 50 cm 50 cm

proton

B A 3 cm

proton

9- In the shown figure, find the value of

(Q2) so that the electric potential at point

(P) equals to (zero) and calculate also the

total electric potential energy for the

whole system of charges. 10- Two conducting spheres are separated from each other by a long distance. The first has

a radius (20 cm) and the potential of its surface equals (270 volt), while the second has a radius (60 cm) and the electric potential of its surface equals (180 volt). The spheres are connected together by a conducting wire. Find: a- The charge per unit area for each sphere after connection. b- The quantity of charge transferred between the spheres after connection.

11- A non-conducting sphere with radius (R=5 cm) is charged uniformly so that the electric

field intensity is given as:

Find the electric potential at a point with distance (r) from the sphere center, if (r =0.0, 3.0 and 8.0 cm).

12- A wire (rod) having a uniform linear charge density (= 4 nC/m) is bent into the shape shown in Figure. Find the electric potential at point O. (Take R=5 cm)

13- A uniformly charged ring has a radius (8 cm) and a total charge (6 nC). When an

electron is projected with velocity (2×107 m/s) from point (A) towards point (B), find its velocity when reaching point (B). Points (A, B) are on the axis of the ring with distance (10, 25 cm) from ring center, respectively.

14- A conducting sphere has a radius (7 cm) and a total charge (4 nC). Find the electric field and the electric potential at a point with distance (r) from the sphere center, if (r =0.0, 5.0 and 9.0 cm).

Q4 = 2 nC

Q1 = 4 nC Q2

Q3 = -8 nC

P

10 cm

6 cm

cmc

mcm

Answers

1- 4.35×105 m/s. 2- -1235 volt, 1.42×10-6 J. 3- 0.766 J.

4- E = 0 r a, E =

b r a, E =

r b.

V = 738 r a, V = 450 +

b r a, V =

r b

5- VA = 13500 volt, VB = -13500 volt, velocity (at point A) = 1.956×106 m/s. and in the second case velocity at point (B) = 3.76×106 m/s.

6- a- the point is (1.6 m) from the positive charge and (2.4 m) from the

negative charge, V = -11250 volt. b- the point lies between the two charges x = 0.554 m from the negative charge, E= 8.58×105 N/C.

7- a- 1 = 0.66 C/m2 , 2 = 0.398 C/m2 .

b- Qtransferred = 5.25 C.

c- (V)1 = -157500 volt, (V)2 = 94500 volt. 8- VB – VA = -16.97 volt, velocity at (B) = 5.7×104 m/s. 9- Q2 = 2 nC, U = -4.08×10-6 J.

10- 1 = 8.95 nC/m2 , 2 = 2.98 nC/m2 , Q (trans.) = 1.5 nC. 11- At (r= 0) V=1080 volt, At (r=3 cm) V=950.4 volt, at (r= 8 cm) V=450 volt. 12- V=192.2 volt. 13- Velocity at B ( vB = 1.8×107 m/s). 14- At r=0.0 cm E=0.0, V=514.3 volt

At r=5.0 cm E= 0.0, V=514.3 volt At r=9.0 cm E= 4444.4 N/C, V= 400 volt

Important Reading Problem - An insulated charged sphere has a radius of (R1=5 cm) and a total charge

(Q1=4 pC). Concentric with this sphere, there is a hollow conducting charged sphere with internal and external radius (R2=7 cm, R3=9 cm) and has a charge (Q2=3 pC). Find the electric field intensity and the electric potential at points (1, 2, 3 and 4) if (r1=10, r2=8, r3=6 and r4=3 cm).

Answer

First we will find the electric field intensity in different regions:

For point (4), rR1: Take gauss surface as sphere with radius r so that:

Q occupies a volume of (

Q1(4 pC) occupies a volume (

So

So

rR1

For point (3), R1 r R2: Take gauss surface as sphere so that:

Since the total charge inside Gauss surface equals to (Q1= 4 pC). Here we have

For point (2), R2 r R3:

Q2=3 pC

-4 pC

7 pC

2 3 4

R1 R2

R3

1 A B

C

Q1=4 pC

E = 0 R2 r R3 Because this region is inside the hollow conducting sphere and the field is zero.

For point (1), r R3: The total charge inside Gauss surface is the total charge of the two spheres, it equals (7 pC) so:

and

To get the electric potential at point (1)

Point (A) can be considered outside the hollow sphere so

Points (A,2 and B) all can be considered inside the hollow conducting sphere, for this reason they have the same potential

To get V3

We can consider point (C) outside the solid insulating sphere so that Vc can be obtained from V3 by substituting for r with (R1)

To get V4

so that

E at point (1) = 6.3 N/C E at point (2) = 0 E at point (3) = 10 N/C E at point (4) = 8.64 N/C V at point (1) = 0.63 volt V at point (2) = 0.7 volt V at point (3) = 0.79 volt V at point (4) = 1.14 volt