When the plates of a capacitor are pushed closer together, the capacitance of the capacitor increases. This is because the capacitance (C) of a parallel plate capacitor is directly proportional to the area (A) of one of the plates and inversely proportional to the distance (d) between the plates (C=εA/d, where ε is the permittivity of the medium between the plates).

Since the capacitor is disconnected from the circuit, the charge (Q) on the capacitor remains constant. However, because the capacitance (C) increases when the plates are pushed closer together, the voltage (V) across the capacitor decreases (since V=Q/C).

The electric field (E) between the plates of a capacitor is given by E=V/d. Since both the voltage (V) and the distance (d) decrease, the effect on the electric field depends on how much each quantity changes. If the voltage decreases more than the distance, the electric field will decrease. If the distance decreases more than the voltage, the electric field will increase.

So, the statement that is true about the electric field and the voltage difference between the plates when they are pushed closer together is: The voltage difference decreases, and the change in the electric field depends on the relative change in the voltage and the distance.

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When the plates of a capacitor are pushed closer together, the capacitance of the capacitor increases. This is because the capacitance (C) of a parallel plate capacitor is directly proportional to the area (A) of one of the plates and inversely proportional to the distance (d) between the plates (C=εA/d, where ε is the permittivity of the medium between the plates).

Since the capacitor is disconnected from the circuit, the charge (Q) on the capacitor remains constant. However, because the capacitance (C) increases when the plates are pushed closer together, the voltage (V) across the capacitor decreases (since V=Q/C).

The electric field (E) between the plates of a capacitor is given by E=V/d. Since both the voltage (V) and the distance (d) decrease, the effect on the electric field depends on how much each quantity changes. If the voltage decreases more than the distance, the electric field will decrease. If the distance decreases more than the voltage, the electric field will increase.

So, the statement that is true about the electric field and the voltage difference between the plates when they are pushed closer together is: The voltage difference decreases, and the change in the electric field depends on the relative change in the voltage and the distance.

Knowee AI · Accepted Answer

When the plates of a capacitor are pushed closer together, the capacitance of the capacitor increases. This is because the capacitance (C) of a parallel plate capacitor is directly proportional to the area (A) of one of the plates and inversely proportional to the distance (d) between the plates (C=εA/d, where ε is the permittivity of the medium between the plates).

Since the capacitor is disconnected from the circuit, the charge (Q) on the capacitor remains constant. However, because the capacitance (C) increases when the plates are pushed closer together, the voltage (V) across the capacitor decreases (since V=Q/C).

The electric field (E) between the plates of a capacitor is given by E=V/d. Since both the voltage (V) and the distance (d) decrease, the effect on the electric field depends on how much each quantity changes. If the voltage decreases more than the distance, the electric field will decrease. If the distance decreases more than the voltage, the electric field will increase.

So, the statement that is true about the electric field and the voltage difference between the plates when they are pushed closer together is: The voltage difference decreases, and the change in the electric field depends on the relative change in the voltage and the distance.

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