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Mitchell
Mitchell
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The capacitor stores its energy in form of electric field.

Initially both the plates were neutral. After applying the potential difference, electrons move form one plate to another.

So to calculate the energy, you would first take a charge $dq$ from one plate to another. By this action, you just created an opposite charge $-dq$ on the other plate the moment you took the charge from it. This will set up an E. field. You set up an electric field that wasn't there before, so some work has to be done.

We will keep taking charge from one plate and deposit it on the other plate till the charge on the plate becomes $Q$.

Let's assume that at a point of time, charge on a plate is $Q'$. Therefore at the same time, the electric field between the plates will be $E=\frac{Q'}{A\epsilon_{\circ}}$.

Now, we will move charge $dq$ and the work done will be given by :

$dW=Ed.dq$

$W=\frac{d}{A\epsilon_{\circ}}\int_0^Q Q'dq$

$W=\frac{1}{C}.\frac{Q^2}{2}$

On rearranging,

$W=\frac{CV^2}{2}$

When you place a $1^{st}$ charge at point A (as per your question) you bring it to such a point where it interacts will no other charge. Thus, we come to the conclusion that work done here is $0$. The work will only be done in bringing the $2^{nd}$ charge at point B.

The capacitor stores its energy in form of electric field.

Initially both the plates were neutral. After applying the potential difference, electrons move form one plate to another.

So to calculate the energy, you would first take a charge $dq$ from one plate to another. By this action, you just created an opposite charge $-dq$ on the other plate the moment you took the charge from it. This will set up an E. field.

We will keep taking charge from one plate and deposit it on the other plate till the charge on the plate becomes $Q$.

Let's assume that at a point of time, charge on a plate is $Q'$. Therefore at the same time, the electric field between the plates will be $E=\frac{Q'}{A\epsilon_{\circ}}$.

Now, we will move charge $dq$ and the work done will be given by :

$dW=Ed.dq$

$W=\frac{d}{A\epsilon_{\circ}}\int_0^Q Q'dq$

$W=\frac{1}{C}.\frac{Q^2}{2}$

On rearranging,

$W=\frac{CV^2}{2}$

When you place a $1^{st}$ charge at point A (as per your question) you bring it to such a point where it interacts will no other charge. Thus, we come to the conclusion that work done here is $0$. The work will only be done in bringing the $2^{nd}$ charge at point B.

The capacitor stores its energy in form of electric field.

Initially both the plates were neutral. After applying the potential difference, electrons move form one plate to another.

So to calculate the energy, you would first take a charge $dq$ from one plate to another. By this action, you just created an opposite charge $-dq$ on the other plate the moment you took the charge from it. This will set up an E. field. You set up an electric field that wasn't there before, so some work has to be done.

We will keep taking charge from one plate and deposit it on the other plate till the charge on the plate becomes $Q$.

Let's assume that at a point of time, charge on a plate is $Q'$. Therefore at the same time, the electric field between the plates will be $E=\frac{Q'}{A\epsilon_{\circ}}$.

Now, we will move charge $dq$ and the work done will be given by :

$dW=Ed.dq$

$W=\frac{d}{A\epsilon_{\circ}}\int_0^Q Q'dq$

$W=\frac{1}{C}.\frac{Q^2}{2}$

On rearranging,

$W=\frac{CV^2}{2}$

When you place a $1^{st}$ charge at point A (as per your question) you bring it to such a point where it interacts will no other charge. Thus, we come to the conclusion that work done here is $0$. The work will only be done in bringing the $2^{nd}$ charge at point B.

Source Link
Mitchell
Mitchell
  • 4.9k
  • 4
  • 25
  • 38

The capacitor stores its energy in form of electric field.

Initially both the plates were neutral. After applying the potential difference, electrons move form one plate to another.

So to calculate the energy, you would first take a charge $dq$ from one plate to another. By this action, you just created an opposite charge $-dq$ on the other plate the moment you took the charge from it. This will set up an E. field.

We will keep taking charge from one plate and deposit it on the other plate till the charge on the plate becomes $Q$.

Let's assume that at a point of time, charge on a plate is $Q'$. Therefore at the same time, the electric field between the plates will be $E=\frac{Q'}{A\epsilon_{\circ}}$.

Now, we will move charge $dq$ and the work done will be given by :

$dW=Ed.dq$

$W=\frac{d}{A\epsilon_{\circ}}\int_0^Q Q'dq$

$W=\frac{1}{C}.\frac{Q^2}{2}$

On rearranging,

$W=\frac{CV^2}{2}$

When you place a $1^{st}$ charge at point A (as per your question) you bring it to such a point where it interacts will no other charge. Thus, we come to the conclusion that work done here is $0$. The work will only be done in bringing the $2^{nd}$ charge at point B.