(a) In Figure 24-33, what is the net electric potential (voltage) at point P due

Question

(a) In Figure 24-33, what is the net electric potential (voltage) at point P due to the four fixed particles if V = 0 at infinity, q = 705 nC, and d = 3.50 cm? (b) Suppose we move a fifth particle that has charge 236 nC starting from point P and ending up at infinity moving to the left along a line extending through the two negative charges. What would be the change in the electric potential energy for this fifth particle? (c) If a different path were taken between P and infinity, the potential energy change would be smaller would be the same could be larger or smaller depending on the path. would be larger

Explanation / Answer

Electric potential is given as ::

V = kq/r

Total Electric potential at P is given as

V = kq/d + kq/d + k(-q) /d + k (-q) /(2d)

V = (kq/d) - (kq/2d)

V = kq/2d

V = (9 x 109) (705 x 10-9) / (2 x 3.50 x 10-2)

V = 9.06 x 104 volts

b)

q' = new charge = 236 x 10-9 C

Change in electric Potential energy = q' V = (236 x 10-9 ) (9.06 x 104) = 2138.16 x 10-5 = 0.02138 J

c)

would be same since electric Potential energy does not depend on path followed

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