Question

Two parallel plate capacitor with diff distance between the plates are connected in parallel to a voltage source. A point positive charge is moved from a point 1 that is exactly in middle between the plates of capacitor C1 to a point 2 in capacitor C2 that lies at a distance from the negative plate of C2 equal to half the distance between the plates of C1.Then the work done by electric force is

Answer 1

@ganeshie8

Answer 2

Options are
A).positive
B).negative
C).0
D).data sufficient

Answer 3

Oops insufficient

Answer 4

any picture given ?

Answer 5

Yep !

Answer 6

Bt i don't have option to draw...

Answer 7

you may attach the file..

Answer 8

No such option here ..
Only post is written...

Answer 9

you're on mobile ?

Answer 10

Yep..

Answer 11

tell me the name of your textbook

Answer 12

Actually the diagram is just depicting the situation given in the question and nothing else..

Answer 13

R u able to draw the diagram requested in the question???

Answer 14

the situation of the problem can be represented by this drawing:

so the work done by the electric field, is:
\[{L_E} = \left( {4\pi {\sigma _1}\frac{{{d_1}}}{2} - 4\pi {\sigma _2}\frac{{{d_1}}}{2}} \right)q = 2\pi {d_1}\left( {{\sigma _1} - {\sigma _2}} \right)q\]
where \(q>0\), \(d_1,\;d_2\) are the separation of the capacitor \(C_1,\;C_2\) respectively
\(V\) is the external voltage
Next, we can write:
\[{\sigma _1} - {\sigma _2} = \frac{{\left( {{\sigma _1} - {\sigma _2}} \right)S}}{S} = \frac{{\left( {{Q_1} - {Q_2}} \right)}}{S} = \frac{{\left( {{C_1} - {C_2}} \right)V}}{S}\]
furthermore, we have:
\[2\pi {\sigma _1} = \frac{S}{{2{C_1}}}\]
so, after a substitution, we get:
\[\begin{gathered}
{L_E} = \left( {4\pi {\sigma _1}\frac{{{d_1}}}{2} - 4\pi {\sigma _2}\frac{{{d_1}}}{2}} \right)q = 2\pi {d_1}\left( {{\sigma _1} - {\sigma _2}} \right)q \hfill \\
\hfill \\
= \frac{S}{{2{C_1}}} \cdot \frac{{\left( {{C_1} - {C_2}} \right)V}}{S} \cdot q = \boxed{\frac{{\left( {{C_1} - {C_2}} \right)qV}}{{2{C_1}}}} \hfill \\
\end{gathered} \]

Answer 15

How do we have this ?
S/2C1=2πsigma1

Answer 16

And sir you considered just the potential at both points due to one plate of cPacitor and not due to other..
Why so??