Question

A charge qı = 2.0 uC is located at the origin and a charge 92 = -6.0 uC is located on the y

axis, 3.0 m from the origin. (a) Find the total clectric potential due to these charges at the point P, which
is on the x-axis and 4.0 m away from the origin. (b) Find the change in the potential energy of a 3.00 uC
charge as it moves from infinity to point P.

Answer 1

(a) -6300 V

To find the total potential at point P, we can just sum the potential due to the charge q1 and the potential due to the charge q2.

The formula for the potential is:

`V=k(q)/(r)`

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

For charge q1,

`q_1 = 2.0 \mu C = 2.0 \cdot 10^(-6) C`

r = 4.0 m

`V_1(P)=(9\cdot 10^9) (2.0\cdot 10^(-6))/(4.0)=4500 V`

For charge q2,

`q_2 = -6.0 \mu C = -6.0 \cdot 10^(-6)C`

while the distance is:

`r=√((4.0)^2+(3.0)^2)=5.0m`

`V_2(P)=(9\cdot 10^9) (-6.0\cdot 10^(-6))/(5.0)=-10800 V`

So, the total potential at P is

`V(P)=V_1(P)+V_2(P)=4500-10800=-6300 V`

(b) -0.0189 J

The change in potential energy of a charge moving in an electric field is

`\Delta U = q \Delta V` (1)

where

q is the charge

`\Delta V` is the potential difference

Here the charge starts from infinity, where the potential is

`V(\infty)=0`

And the charge is

`q=3.00 \mu C = 3.00\cdot 10^(-6)C`

And solving for eq.(1), we find

`\Delta U =q(V(P)-V(\infty))= (3.00\cdot 10^(-6))(-6300 V-0)=-0.0189 J`