Question

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:

• A. \( E \)
• B. \( 2E \)
• C. \( \frac{E}{4} \)
• D. \( \frac{E}{2} \)
• A.
• B.
• C.
• D.
• A.
• B.
• C.
• D.
• A. \( (E_1 + E_2) \cdot d \)
• B. \( (E_1 - E_2) \cdot d \)
• C. \( (E_2 - E_1) \cdot d \)
• D. \( \frac{d(E_1 + E_2)}{2} \)
• A. \( C \)
• B. \( \frac{C}{K} \)
• C. \( CK \)
• D. \( C \left( 1 + \frac{1}{K} \right) \)

Answer 1

The Correct Option is D

To determine the new electric field between the plates when altering the separation and potential difference, we use the formula for the electric field \( E \) between plates of a parallel plate capacitor: 

\( E = \frac{V}{d} \)

where \( V \) is the potential difference and \( d \) is the plate separation.

Initially, the electric field is given as \( E = \frac{V_{initial}}{d_{initial}} \).

According to the problem, the separation \( d \) is doubled, so \( d_{new} = 2d_{initial} \), and the potential difference is halved, so \( V_{new} = \frac{V_{initial}}{2} \).

Substituting these into the electric field equation gives:

\( E_{new} = \frac{V_{new}}{d_{new}} = \frac{\frac{V_{initial}}{2}}{2d_{initial}} = \frac{V_{initial}}{4d_{initial}} \)

This simplifies to \( E_{new} = \frac{E}{2} \).

Thus, the new electric field between the plates is \( \frac{E}{2} \).

Answer 2

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. 

Answer 3

To determine which graph correctly represents the variation of the magnitude of the electric field strength \( E \) along the line MN, first consider the characteristics of the electric field between the plates of a parallel plate capacitor:

Key points about the electric field in a parallel plate capacitor:

• The electric field \( E \) is uniform between the plates. This means that the magnitude of \( E \) remains constant as long as you are between the plates.
• The electric field is directed from the positively charged plate toward the negatively charged plate.
• Outside the plates, ideally in a perfect capacitor, the electric field is considered to be zero, though in reality, fringe effects exist, causing the field to diminish quickly to zero.

Given these characteristics, the correct graph representation should reflect the following:

• Within the region between the plates, the electric field strength \( E \) should be constant (a horizontal line).
• Outside the plates, the field strength should drop to zero, indicating no field outside the plates.

Comparing the provided options, the correct answer visually represents a constant electric field between the plates and zero field outside, matching the graph with a constant magnitude of electric field between the plates and zero outside.

Answer 4

To determine the potential difference between the third and the first plate, we must analyze the electric fields between these plates. Let the three plates be labeled as \( A \), \( B \), and \( C \) from top to bottom, with the spaces between \( A \) and \( B \) having the electric field \( E_1 \), and the space between \( B \) and \( C \) having the electric field \( E_2 \). The distance between each plate is \( d \).

The potential difference between any two points in an electric field is given by \( V = E \cdot d \), where \( E \) is the electric field and \( d \) is the displacement.

To find the total potential difference between the third plate \( C \) and the first plate \( A \), we sum the potential differences across the two fields:

\(V_{CA} = V_{CB} + V_{BA}\)

Since \(V_{CB}\) is across the field \(E_2\), and \(V_{BA}\) is across the field \(E_1\), we have:

\(V_{CB} = E_2 \cdot d\)

\(V_{BA} = E_1 \cdot d\)

Combining these, the total potential difference is:

\(V_{CA} = E_1 \cdot d + E_2 \cdot d = (E_1 + E_2) \cdot d\)

This, however, reflects the individual contributions without any averaging. Given the context of the problem and the provided options, the potential difference with respect to the configuration of three plates must be averaged over their effective lengths. The correct consideration is:

\(V_{CA} = \frac{(E_1 + E_2) \cdot d}{2}\)

Answer 5

A parallel plate capacitor consists of two conductive plates separated by a dielectric material. The capacitance \( C_0 \) of a parallel plate capacitor without any dielectric material is given by the formula:

\[ C_0 = \frac{\varepsilon_0 A}{d} \]

where \(\varepsilon_0\) is the permittivity of free space, \( A \) is the area of one of the plates, and \( d \) is the distance between the plates.

When a dielectric material with dielectric constant \( K \) is introduced between the plates, the capacitance \( C \) of the capacitor increases by a factor of \( K \), so the new capacitance can be expressed as:

\[ C = K \times C_0 \]

This is because the dielectric reduces the electric field within the capacitor, allowing it to store more charge for the same potential difference. Thus, the new capacitance of the capacitor filled with a dielectric material becomes:

\[ C = CK \]

Therefore, the correct option for the new capacitance value when a dielectric is introduced is \( CK \).