A parallel plate capacitor has two parallel plates which are separated by an insulating medium like air, mica, etc. When the plates are connected to the terminals of a battery, they get equal and opposite charges, and an electric field is set up in between them. This electric field between the two plates depends upon the potential difference applied, the separation of the plates and nature of the medium between the plates.
Question: 1
The electric field between the plates of a parallel plate capacitor is \( E \). Now the separation between the plates is doubled and simultaneously the applied potential difference between the plates is reduced to half of its initial value. The new value of the electric field between the plates will be:
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The electric field in a parallel plate capacitor is directly proportional to the potential difference and inversely proportional to the separation between the plates.
The electric field \( E \) between the plates of a parallel plate capacitor is given by:
\[
E = \frac{V}{d}
\]
Where:
- \( V \) is the potential difference across the plates,
- \( d \) is the separation between the plates.
When the separation \( d \) is doubled, and the potential difference \( V \) is reduced to half, the new electric field \( E' \) is given by:
\[
E' = \frac{V/2}{2d} = \frac{E}{2}
\]
Thus, the new electric field between the plates will be half of the initial value, corresponding to option (D).
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Question: 2
A constant electric field is to be maintained between the two plates of a capacitor whose separation \( d \) changes with time. Which of the graphs correctly depict the potential difference (V) to be applied between the plates as a function of separation between the plates (\( d \)) to maintain the constant electric field?
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To maintain a constant electric field in a parallel plate capacitor, the potential difference must be proportional to the separation between the plates.
The electric field \( E \) between the plates of a parallel plate capacitor is related to the potential difference and separation by:
\[
E = \frac{V}{d}
\]
To maintain a constant electric field, the potential difference \( V \) must be directly proportional to the separation \( d \). Therefore:
\[
V = E \cdot d
\]
This relationship indicates that \( V \) increases linearly with \( d \). Hence, the graph of \( V \) versus \( d \) will be a straight line, confirming that option is correct.
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Question: 3
In the below figure, P and Q are the two parallel plates of a capacitor. Plate Q is at positive potential with respect to plate P. MN is an imaginary line drawn perpendicular to the plates. Which of the graphs shows correctly the variations of the magnitude of electric field strength \( E \) along the line MN?
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The electric field between two parallel plates of a capacitor is uniform, and it is zero outside the plates.
The electric field between two parallel plates of a capacitor is uniform and directed from the positive to the negative plate. Between the plates, the electric field is constant. Outside the plates, the electric field is zero.
In the given diagram, plate Q is at a positive potential, and plate P is at a negative potential. The electric field is directed from plate Q to plate P. Along the line MN, which is perpendicular to the plates, the electric field strength will be uniform between the plates and zero outside.
Thus, the correct graph showing the electric field strength variation would be a constant value between the plates, and zero outside the plates. This corresponds to option .
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Question: 4
Three parallel plates are placed above each other with equal displacement \( d \) between neighbouring plates. The electric field between the first pair of the plates is \( E_1 \), and the electric field between the second pair of the plates is \( E_2 \). The potential difference between the third and the first plate is:
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For multiple parallel plates, the potential difference between the plates is the sum of the potential differences across each section of the capacitor.
The potential difference between two plates is given by the product of the electric field and the separation between the plates.
If \( E_1 \) is the electric field between the first pair of plates and \( E_2 \) is the electric field between the second pair of plates, then the potential difference between the first and third plates is the sum of the individual potential differences across the two sections.
The potential difference between the plates is:
\[
V = E_1 \cdot \frac{d}{2} + E_2 \cdot \frac{d}{2} = \frac{d(E_1 + E_2)}{2}
\]
Thus, the correct answer is \( \frac{d(E_1 + E_2)}{2} \), corresponding to option (D).
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Question: 5
A material of dielectric constant \( K \) is filled in a parallel plate capacitor of capacitance \( C \). The new value of its capacitance becomes:
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The capacitance of a parallel plate capacitor increases by a factor equal to the dielectric constant when a dielectric material is inserted.
When a dielectric material of dielectric constant \( K \) is inserted into a parallel plate capacitor, the capacitance increases by a factor of \( K \). The new capacitance \( C' \) is given by:
\[
C' = K \cdot C
\]
Thus, the new capacitance becomes \( CK \), corresponding to option .
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