i dont understand how to do this Q +10 HC 50 cm A charge of +10 HC sits 10 cm be

Question

i dont understand how to do this

Q +10 HC 50 cm A charge of +10 HC sits 10 cm below the top plate of a capacitor. The voltage between the top and bottom plate is 100 V. Problems: 1) What is the electric field at the point charge? 2. What is the electric field at the bottom plate? 3. What is the voltage difference between the 10 HC charge and the bottom place of the capacitor? 4. When the charge Q moves from where it is 10 cm below the top plate to the bottom plate what energy will it have gained

Explanation / Answer

1) E = V/d = 100/0.5 = 200 N/C

2) Same ; Field is constant inside the capacitor.

3) V at charge = E*distance between the points considered = 200 * 0.4 = 80 V

4) Energy gained = Potential difference * Charge = 80*10*10^-6 = 0.8 mJ

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