Question

A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in it

A. Zero
B. 15 joule
C. 45 joule
D. 1500 joule​

Answer 1

A. Zero

Step-by-step 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,

`k=(1)/(4\pi\epsilon_(0)  )`

= 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