Question

CHEMISTRY
Question No. 120
Duration: 3:00:00 Time Left: 0
Two charges 1 micro coulomb and 4 micro coulombs are separated by 20 centimeters. The work done to increase the separation
between the charges to 40 centimeters very slowly is
(A)
- 0.18 J
(B)
0.18 J
(C)
-0.36 J
(D)
0.36 J
A
B
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Answer 1

Given that,

Charge $q_{1}=1 \mu C$

Other charge = 4 μC

Distance d= 20 cm

We need to calculate the potential

Using formula of potential

$V_{1}=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q_{1}q_{2}}{d_{1}^2}$

Put the value into the formula

$V_{1}=9\times10^{9}\times\dfrac{1\times10^{-6}\times4\times10^{-6}}{(20\times10^{-2})^2}$

$V_{1}=0.9\ V$

We need to calculate the second potential

Using formula of potential

$V_{2}=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q_{1}q_{2}}{d_{2}^2}$

Put the value into the formula

$V_{2}=9\times10^{9}\times\dfrac{1\times10^{-6}\times4\times10^{-6}}{(40\times10^{-2})^2}$

$V_{2}=0.225\ V$

We need to calculate the work done to increase the separation

Using formula of work done

$W=V_{2}-V_{1}$

Put the value into the formula

$W=0.225-0.9$

$W=-0.68\ J$

Hence, The work done to increase the separation is -0.68 J.