Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance moved against the field of intensity, r = 30 cm
= 0.3 m
The electric field, E = 50 N/C
The formula for energy stored in the charge is,
V = k q/r
Where,
[tex]k=\frac{1}{4\pi\epsilon_{0} }[/tex]
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
