An isolated large conducting plate having area A and total charge Q is placed in a uniform electric field →E as shown in the figure. If q1 and q2 are the charges appearing on the two sides of the plate, then
A
q1=Q2−EAε0, q2=Q2+EAε0
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B
q1=Q2+EAε0 , q2=Q2−EAε0
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C
q1=Q3, q2=Q2−EAε0
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D
None of the above
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Solution
The correct option is Aq1=Q2−EAε0, q2=Q2+EAε0 Given,
Net charge on the plate=Q
Area of the plate=A
Let us consider a point P inside the plate as shown in figure.
As we know that, electric field at point P inside the conductor is zero. So, Enet=0
Electric field at point P due to charge q1 is q12Aεo as shown in figure.
Similary,
Electric field at point P due to charge q2 is q22Aεo as shown in figure.
Thus, net electric field at point P,
Enet=q12Aεo+E−q22Aϵ0
Since, Enet=0, so we have
q12Aεo+E−q22Aϵ0=0.......(1)
Also, from conservation of charge, q1+q2=Q.......(2)