Atest 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

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,

                                      = 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

1/0.14nm = 10^10/1.4 a lot