a) What happens to charge when the plate separation is and why? (Please increas

Question

a) What happens to charge when the plate separation is and why? (Please increas take note of any charge flow) b) What happens to the strength of the field when the separation is increased, and 5) With the separation 5 mm, disconnect the battery. a) What happens to charge when the plate separation is increased, and why? b) What happens to the voltage when the plate separation is increased, and why? c) What happens to the strength of the field when plate separation is increased? (You can use the field meter to check this. 6) Repeat experiment 5) a) but now use the stored energy meter. Why do you think the change takes place.

Explanation / Answer

4(a) The potential difference across Capacitor plates is 1.5V, being parallel plate capacitors their capacitance is epsilon_not *A/d , d is separation between plates, and if this separation is increased Capacitance of capacitors will decrease. And the potential is constant i.e 1.5V. Again Q=CV so charge also reduces on plates.
(b) Electric field inside parallel plate capacitor is not dependent on distance from plates, it is constant i.e charge density/epsilon_not. So Strength os electric field is constant after increasing separation.

5(a) Now as battery is removed, and separation is increased, Capacitance will decrease and the potential difference between plates of capacitors increases but charge remains constant.
(b) Potential difference increases as charge must remain constant and Capacitance is reduced.
(c) Again Electric field remains constant as it is constant between the plates for any distance

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