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joja [24]
3 years ago
6

An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then di

sconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Physics
1 answer:
DedPeter [7]3 years ago
3 0

Answer:

The new voltage between the plates of the capacitor is 18 V

Explanation:

The charge on parallel plate capacitor is calculated as;

q = CV

Where;

V is the battery voltage

C is the capacitance of the capacitor, calculated as;

C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}

q = \frac{\epsilon _0A V}{d}

where;

ε₀ is permittivity of free space

A is the area of the capacitor

d is the space between the parallel plate capacitors

If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d}  = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1}  = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V

Therefore, the new voltage between the plates of the capacitor is 18 V

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