Question

Let \(E_1\), \(E_2\), \(E_3\), and \(E_4\) be the magnitudes of the electric field between the pairs of plates, I, II, III, and IV respectively. Then:

• A. \(E_1 > E_2 >E_3 >E_4\)
• B. \(E_3 > E_4 >E_1 >E_2\)
• C. \(E_4 > E_3 >E_2 >E_1\)
• D. \(E_2 > E_3 >E_4 >E_1\)

Hint:

Always remember, the electric field strength is proportional to the voltage difference divided by the separation distance. Greater voltage differences over the same distance result in stronger fields.

Answer 1

The Correct Option is C

Given that the electric field \( E \) between two plates is calculated as:

\[ E = \frac{\Delta V}{d} \]

where \( \Delta V \) is the potential difference and \( d \) is the distance between the plates (0.02 m in this case).

Pair I:

\[ \Delta V = -70V - (-50V) = -20V \] \[ E_1 = \frac{20}{0.02} = 1000 \, V/m} \]

Pair II:

\[ \Delta V = 150V - (-50V) = 200V \] \[ E_2 = \frac{200}{0.02} = 10000 \, V/m} \]

Pair III:

\[ \Delta V = 200V - (-20V) = 220V \] \[ E_3 = \frac{220}{0.02} = 11000 \, V/m} \]

Pair IV:

\[ \Delta V = -100V - (-400V) = 300V \] \[ E_4 = \frac{300}{0.02} = 15000 \, V/m} \]

Answer:

Comparing the magnitudes:

\[ E_4>E_3>E_2>E_1 \]

Thus, the correct option is \( (C)} \, E_4>E_3>E_2>E_1 \).

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