Question

Two charges lie on the x-axis, one at the origin and the other at 14.3m What is the potential (relative to infinity) due to these charges at a point at 7.1m measured from the origin on the x-axis?

Answer 1

V = 2.52 10³ V

Step-by-step explanation:

The electric potential for a point charge is

V = k ∑
`(q_i)/(r_i)`

In this case the formula is

V = k (
`(q_1)/(r_1) + (q_2)/(r_2)`)

distances are the absolute value

r₁ =√ (7.1 -0)² = 7.1 m

r₁ =√ (7.1 - 14.3)² = 7.2 m

we substitute

V = 9 10⁹ (q₁ / 7.1 + q₂ / 7.2)

we have two possibilities

* different charges

V = 9 10⁹ (q₁ / 7.1 + q₂ / 7.2)

* equal charges and same sign

q₁ = q₂ = q

V = 9 10⁹ q (1 / 7.1 + 1 / 7.2) = p 9 10⁹ 0.2797

V = 2.52 10⁹ q

if we assume a value of the charge, for example q = 1 10⁻⁶ c

V = 2.52 10⁹ 1 10⁻⁶

V = 2.52 10³ V