A,B and C are three identical, infinitely long , parallel conducting plates placed as shown in the figure. If q1 and q2 be the charges present on the surfaces left of A and right of B then find the value of q1 and q2
A
q1=3Q2, q2=−Q2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
q1=−3Q2, q2=Q2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
q1=3Q2, q2=Q2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
q1=3Q2, q2=3Q2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Cq1=3Q2, q2=Q2 Let conducting plate A has charge"−Q" , conducting plate B has charge "3Q" and conducting plate C has charge "Q".
Let us assume the charge distribution on all the plates is as shown in the figure and P be a point inside the the conducting plate A.
To find charges q1 and q2
Net field at point P due to charges on each surface:
EP=(−Q−x)2Aε0−(x2Aε0+3Q2Aε0+Q2Aε0)
Since, P is inside the conductor. EP=0
⇒(−Q−x)2Aε0−(x2Aε0+3Q2Aε0+Q2Aε0)=0
⇒x=−5Q2
So, (−Q−x)=(−Q+5Q2)
∴q1=(−Q−x)=3Q2
Since, here electric field inside the Gaussian surface is zero , the net charge enclosed by the Gaussian surface must be zero. So,
y=−x
∴y=−(−5Q2)=5Q2
Now, from conservation of charge, q2+y=3Q
⇒q2=3Q−y
Substituting the value of y we get,
q2=3Q−5Q2
⇒q2=Q2
Hence option (c) is the correct answer.
Why this question?
Key Point: The net electric field at any point is independent of charge distribution on the plate, it only depends on the total charge on a plate.