A parallel-plate capacitor is charged and then kept connected with a 5V battery.

Question

A parallel-plate capacitor is charged and then kept connected with a 5V battery. Then, the plate s decreased by a factor of 4, d_new-1/4_d old, please type in a number into the box to determine how everything changes. (If something doesn't change, please type in 1 to answer X_new- 1 x_old.) (a) By wh fraction does the capacitance change? C_new= 4 C_old; (b) By what fraction does the amount of charge change? Q_new= Q_old [When the battery is connected, charges can keep moving and the charge may not stay to be the same.) (c) By what fraction does the voltage difference between the two plates change? delta V_new- 1/4 delta V_old (When the battery is always connected, the voltage difference across the capacitor will be always equal to the battery voltage.) (d) By what fraction does the energy stored in the capacitor change? U_new= U_old Does It make sense? Let's check, considering the amount of change and the area of the plate, (e) by what fraction does the amount of change density change? o_new= o_old (f) Consider how E is related to the change density, by what fraction does the E field change? E_new= 1/4 E_old. Considering how E, d and voltage difference are related, does it make sense now, how E, and d change to keep the voltage difference unchanged? Pease explain why the energy stored in the capacitor changes? If Increased, where did the energy come from? If it decreased, where could the energy go?

Explanation / Answer

According to the given problem,

Using the formula for,

C = 0A/d whrer d os separation

C = q/V and E = 1/2CV2

a) Cnew = 4Cold

b) Qnew = 4Q

c) Vnew = Vold = 5V

d) Enew = 4Eold

Only 4 Subparts per question, Repost the remaing separatly.

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