Question

a parallel plate capacitor has 1microfarad capacitance . one of its two plates is given 2micro coulomb charge and the other has 4 micro coulomb cahrge. the potential difference developed across the capacitor is
3v
1v
5v
2v

Answer 1

The potential difference is 5 V.

Explanation:

q1 = 1 μC = 1 x 10^-6 C

q2 = 2μC = 2 x 1-^-6 C

C = 0.1 μF = 1 x 10^-7 F

Net q = q1 - q2 / 2  = ( 1 - 2) x 10^-6 / 2 = -0.5 x 10^-6 C

Potential V = q / C

Potential  "V" = 1 x 10^-7 / - 5 x 10^-7 = - 5V

Potential difference can never be (-) therefore it is 5 V.

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Three capacitors of capacitance 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.?

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Answer 2

Let surface charge density be denoted by *

Let epsilon nought be denoted by @

E = */2@

net E = (*1-*2)/2@

* = q/A

Net E = (q1-q2)/2A@

Given charges are 4 uC & 2 uC

Let q1 = 4 uC & q2 = 2 uC

Net E = 1/A@

V = Ed

V = d/A@

C = A@/d

V = 1/C

Given capacitance is 1 uF

V = 1/1

Potential difference = 1 V

Option 2 is correct

:-)