Question

For the isolated charged conductor shown in the
figure, if the electric field at point A and B is E
and E, respectively, then

+
+
B
+ + + + +
+
A
+
+
+
+
+
+
(1) E. > EA
(3) EA = Eg+ 0
(2) EA > EB
(4) EA = Ep = 0
Alostri
JH​

Answer 1

Answer:

E a>E b as radius of curvature near Pt.A is smaller than that of pt. B

Answer 2

For the Isolated charged conductor the option, Ea=Eb=0 is the correct answer.

• To find the answer we need to find the electric field inside the conductor.  For this conductor, we have positive and negative charges.
• Let, Ext be an external electric force being applied. With this external force on the conductor, the negative charges will move in the force direction and negative charges will move in the opposite direction.
• These positive charges will cancel the external charge. And now the electric field inside the conductor will be zero.
• $E=-dV/dx$$=0$, so V is constant. Let V₀ be the electric potential. V₀ value will be constant throughout the surface of the conductor.
• At point A, the potential will be V₀ = Vd.
• As the potential is constant then Ea=Eb=0

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