\(\frac{E^2Ad}{\epsilon_0}\)
\(\frac{1}{2}\epsilon_0E^2\)
\(\epsilon_0EAd\)
\(\frac{1}{2}\epsilon_0E^2Ad\)
To find the energy stored in a parallel plate capacitor with a uniform electric field \(\textbf{E}\) between the plates, we need to understand the relationship between the electric field, the charge, and the potential difference.
The relevant parameters given are:
Step 1: Understand the potential difference.
The potential difference \(V\) between the plates is related to the electric field \(E\) and the distance \(d\) by:
\(V = E \cdot d\)
Step 2: Calculate the charge on the capacitor.
The capacitance \(C\) of a parallel plate capacitor is given by the formula:
\(C = \frac{\epsilon_0 \cdot A}{d}\)
Therefore, the charge \(Q\) on the capacitor can be related to the capacitance and potential difference by:
\(Q = C \cdot V = \frac{\epsilon_0 \cdot A}{d} \cdot E \cdot d = \epsilon_0 \cdot A \cdot E\)
Step 3: Calculate the energy stored in the capacitor.
The energy \(U\) stored in a capacitor is given by:
\(U = \frac{1}{2} \cdot C \cdot V^2\)
Substitute the expressions for \(C\) and \(V\):
\(U = \frac{1}{2} \cdot \frac{\epsilon_0 \cdot A}{d} \cdot (E \cdot d)^2\)
\(= \frac{1}{2} \cdot \epsilon_0 \cdot A \cdot E^2 \cdot d\)
Thus, the energy stored in the capacitor is:
\(U = \frac{1}{2} \epsilon_0 E^2 A d\)
Conclusion:
The correct option for the energy stored in the capacitor is:
\(\frac{1}{2}\epsilon_0E^2Ad\)
Hence, the correct answer is the fourth option: \(\frac{1}{2}\epsilon_0E^2Ad\).
Find work done in bringing charge q = 3nC from infinity to point A as shown in the figure : 
Two capacitors \( C_1 \) and \( C_2 \) are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are \( U_1 \) and \( U_2 \), respectively. Which of the given statements is true? 
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.
Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor \( \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \), where \( \omega \) is the frequency of the supply across resistor \( R \) and inductor \( L \). If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.
In light of the above statements, choose the most appropriate answer from the options given below:
What is Microalbuminuria ?
The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.
Capacitors commonly known as Condensers are passive components, similar to a resistor. In capacitors, charges are usually stored in the form of an "electrical field". Electrical and electronic circuits depend on the same which is made up of two parallel metal plates that are not connected to one another. The two plates are separated by a non-conducting insulating medium called dielectric.
Read More: Types of Capacitors