Question

When a point charge of 6c is moved between two points In an electric field the work done is 1.8×10^-5j. The potential difference between the points is

Answer 1

Explanation:

we know:

∆V= Work done ( by external agent)/q

so:

∆V= 1.8x10^-5 /6 .

= 3 × 10^-6 J/c.

Thank you.

Answer 2

Given: Charge, Q = 6 C

           Work done, W = 1.8 x 10⁻⁵ J

To Find: Potential difference, ΔV.

Solution:

To calculate ΔV, the formula used:

• Work done = charge x potential difference
• W = Q X ΔV

Applying the above formula-

1.8 x 10⁻⁵ = 6 xΔ V

1.8 x 10⁻⁵ /6 = ΔV

ΔV = 1.8 x 10⁻⁵ /6

  = 0.3 x 10⁻⁵

ΔV = 3 x 10⁻⁶ volt or 3 x 10⁻⁶ J/C

Hence, the potential difference between the two points is 3 x 10⁻⁶ volts.