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A parallel plate capacitor of capacitanc...

A parallel plate capacitor of capacitance `3muF` has total charge `+15muC` on one plate and total charge `-15muC` on the other plate. The separation between the plates is 1mm. The electric field between the plates has magnitude: (in N/C)

A

50

B

100

C

5000

D

10000

Text Solution

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The correct Answer is:
To find the electric field between the plates of a parallel plate capacitor, we can follow these steps: ### Step 1: Identify the given values - Capacitance \( C = 3 \mu F = 3 \times 10^{-6} F \) - Charge on one plate \( Q = +15 \mu C = 15 \times 10^{-6} C \) - Charge on the other plate \( -Q = -15 \mu C \) - Separation between the plates \( d = 1 mm = 1 \times 10^{-3} m \) ### Step 2: Calculate the potential difference (\( \Delta V \)) The potential difference across the capacitor can be calculated using the formula: \[ \Delta V = \frac{Q}{C} \] Substituting the values: \[ \Delta V = \frac{15 \times 10^{-6}}{3 \times 10^{-6}} = 5 V \] ### Step 3: Calculate the electric field (\( E \)) The electric field \( E \) between the plates of a capacitor is given by the formula: \[ E = \frac{\Delta V}{d} \] Substituting the values: \[ E = \frac{5}{1 \times 10^{-3}} = 5000 \, N/C \] ### Final Answer The magnitude of the electric field between the plates is \( 5000 \, N/C \). ---

To find the electric field between the plates of a parallel plate capacitor, we can follow these steps: ### Step 1: Identify the given values - Capacitance \( C = 3 \mu F = 3 \times 10^{-6} F \) - Charge on one plate \( Q = +15 \mu C = 15 \times 10^{-6} C \) - Charge on the other plate \( -Q = -15 \mu C \) - Separation between the plates \( d = 1 mm = 1 \times 10^{-3} m \) ...
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