In part B, WHY does the electric potential difference remain the same!!! Two lar

Question

In part B, WHY does the electric potential difference remain the same!!!

Two large flat plates are separated by a distance d. The plates are connected to a battery. The surface area of the face of each plate is A_1. Write an expression for the capacitance in terms of A_1 and d. A new capacitor is formed by attaching two uncharged metal plates, each with area A_2, to the capacitor as shown. The battery remains connected. When the new plates are attached, does the electric potential difference between the plates increase, decrease, or remain the same? Explain. Write an expression for the work done by the electric field on a charge +q_o as it travels from the left plate of the capacitor to the right. Express your answer in terms of the given variables. Explain. Write an expression for the magnitude and direction of the electric field between the plates. Is the magnitude of the electric field greater than, less than, or equal to the magnitude of the electric field between the plates before the new plates were attached? Write an expression for the charge density on the plates of the capacitor. Is the charge density greater than, less than, or equal to the charge density on the plates before the new plates were attached? Explain.

Explanation / Answer

a) C = A1*epsilon/d

b)

i) remain the same.

because, the plates are connected battery.

the potential diffrnec between the terminal of the battery must be equal to potential diffrnce between the plates.

ii)
Workdone = F*d

= q*E*d

= Vo*d

iii)

E = Vo/d

E remain the same.

the direction of E is towards -x axis (from right plate to left plate)

iv) charge density, sigma = Q/A

= C*Vo/A

= (A*epsilon/d)*Vo/A

= epsilon*Vo/d

charge density density also remain the same.

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