Question

13)A conductor when given a charge of 10 mC acquires a potential of 250 V.The capacity of conductor is a) 0.4uF b) 4uF C) 40uF d)0.25 F​

Answer 1

Answer:

C. 40uF is right

Explanation:

C= Q/V

= 10×10-³ / 250

= 4×10-⁵ F

=40uF

Answer 2

Given : Charge on conductor : Q =10mC

            Potential Difference : V = 250V.

To Find : Capacity of conductor.

Solution :

Charge on conductor : Q =10mC.

Potential Difference : V = 250V.

Thus, Capacitance of Conductor is  

C= $\[\frac{Q}{V}\]$

C= $\[\frac{{10 \times {{10}^{ - 3}}}}{{250}}\]$

C=$\[4 \times {10^{ - 5}}F\]$

C= $\[40 \times {10^{ - 6}}F\]$    

C = 40uF.

Hence, Capacity of conductor is 40uF.