Question

Two identical metal plates are given positive charges Q1 and Q2 (

Answer 1

Answer:

Let A be the area of each metal plate. If they are distance d apart, then the capacitance of the parallel plate capacitor so formed will be

C=∈0Ad

If E1,E2 are electric field intensities due to the two plates (in the space between teh plates), them

E1=q1/A2∈0andE2=q2/A2∈0

Net electric field between the two plates

E=E1−E2=q1/A2∈0−q2/A2∈0=12∈0A(q1−q2)

The potential difference between the plates fo the capacitor, V=E×d=d2∈0A(q1−q2)

V=q1−q22(∈0A/d)=q1−q22C

Explanation:

Answer 2

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