Question
Question: On increasing the plate separation of charged condenser its energy: A. Remains unchanged B. Decr...
On increasing the plate separation of charged condenser its energy:
A. Remains unchanged
B. Decreases
C. Increases
D. None of these.
Solution
here, we will use the formula of the energy of a charged condenser to find the relation between the energy and the distance between the plates. This energy will be dependent on the capacitance and the potential between the plates.
Complete step by step answer:
We will calculate the energy in the charged condenser by using the following formula
U=21CV2
Here, C is the capacitance between the parallel plates and V is the potential between the plates.
Now, capacitance is defined as the ability of an object to store an electric charge. It is also defined as the ratio of the change in an electric field to the change in the electric potential of a system. Therefore, the capacitance between the parallel plates is given by
C=dAε0
Here, A is the area of plates, ε0 is the dielectric constant in free space, and d is the distance between the two plates.
Now, the potential is defined as the ability of a system to do some work. The Potential between the two plates is given by
V=Aε0Qd
Here, Q is the charge on the plates, d is the distance between the plates, A is the area of plates, and ε0 is the dielectric constant in free space.
Now, putting the values of C and V in the formula of energy U , we get
U=21CV2
⇒U=21×dAε0×(Aε0Qd)2
∴U=21×Aε0Q2d
Now, from the above relation, we get
U∝d
Therefore, the energy of the charged condenser will increase with an increase in the separation of the plates.
So, the correct answer is “Option C”.
Note:
As the plates are carrying opposite charges, therefore, the plates will attract each other.
Hence, to separate the plates the mechanical work requires should be more.
This mechanical work is in the form of energy.