Question

A charge of +4.83 nC is at (0, -2.48 m) and a second charge of + 6.35 nC is located at (-3.14 m, 0). What is the electric potential at the origin due to these two charges? Recall that the constant in Coulomb's Law, k_e, has the value of 8.99 X 10^(9) N m^(2)/C^(2).

		17.5 V.
		18.2 V.
		35.7 V.
		3.57 X 10^(10) V.

Answer 1

Electric potential is given by,

V = k*q/r

Where,

r = distance from the particle to the point.

q = charge of the particle.

k = 8.99*10^9 Nm^2/C^2

For charge 1,

q1 = +4.83 nC = 4.83*10^-9 C

r1 = 2.48 m

V1 = kq1/r1 = (8.99*10^9)(4.83*10^-9)/2.48

V1 = 17.50875 V

for charge 2,

q2 = +6.35 nC = 6.35*10^-9 C

r2 = 3.14 m

V2 = k*q2/r2 = (8.99*10^9)(6.35*10^-9)/3.14

V2 = 18.180414 V

Since, electric potential is a scalar quantity.

So, net electric potential = Vnet = V1 + V2

Vnet = 17.50875 + 18.180414 = 35.689164 V

Vnet =35.7 V

So, correct option is C.

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