Question

15. A charge particle of 2uC moves a distance of 10⁴ m against a electric field of 100 N/C. Work done on the charge particle during this movement is
(a) 8J (b) 200 J (c) 2J (d) 100 J

please solve it...​

Answer 1

Given :

Charge of the particle (q) = 2 micro C = $2 * 10^{-6}$

Distance moved (d) = $10^{4}$ m

Electric Field (E) = 100 N/C

To Find :

Work done on the charge particle

Solution :

Electric potential (V) = Ed

                             V  = 100 * $10^{4}$

                             V  =  $10^{6}$ V

Work done on moving a charge particle (W) = Vq

                                                                         = $10^{6} * 2 * 10^{-6}$

                                                                        =  2 J

(c) 2 J is the correct option.

Answer 2

Given:

• Charge is 2uC = 2 × 10^(-6) C.
• Displacement = 10⁴ m
• Electric field intensity = 100 N/C.

To find:

• Work done to move charge ?

Calculation:

First of all, let's calculate the force experienced by charge :

$F = E \times q$

$\implies F = 100 \times (2 \times {10}^{ - 6} )$

$\implies F = 2 \times {10}^{ - 4} \: N$

Now, let's calculate the work done :

$W = F \times d$

$\implies W = (2 \times {10}^{ - 4} ) \times {10}^{4}$

$\implies W = 2 \times {10}^{ - 4 + 4}$

$\implies W = 2 \: joule$

So, work done is 2 Joules (Option c).