Two concentric spherical conducting shells of radii R and r (Rr) are supported by insulating stands. The outer shell is given a charge Qwhile the inner shell is grounded. Find the charge on the inner shellsolnPotential of the inner shell must be zero, since t is grounded.the problem is that potential of inner shell is then why here whole system potential is taken zero
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OptionsPotential of the inner shell must be zero, since t is grounded.
QUESTION : 2
The electric field and the electric potential at a point due to a point charge are 13 V/C and 8 J/C respectively. Find the magnitude of the charge and the distance of the point from the charge.
Options-
8.89 × 1010C, 2m
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4.44 × 1010C, 2m
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8.89 × 1010C, 0.5m
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4.44 × 1010C, 0.5m
QUESTION : 3
Two charges 2×108C and 3×108C are located 30 cm apart. Determine the point from the negative charge at which the electric potential is 0.
Options-
18 cm
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12 cm
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16 cm
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15 cm
Potential at P
⇒ x= 0.18 m
QUESTION : 4Sixty four water droplets, each of radius 1 mm and charge 1010C combine to form a single large drop. Determine the potential of the bigger drop.
Options-
256 V
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126 V
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280 V
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144 V
Volume of large drop = Volume of small drops
QUESTION : 5
Considera coordinate axis system. At origin, the potential is zero. In the positive x direction, an electric field of intensity E0is directed. Find the potential at x = +x0.
Options
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Ex0
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Ex0
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2Ex0
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2Ex0
For a uniform fieldΔV = EΔr. In this case,ΔV = V 0,Δr = x0 0, and E = E0.
Thus,
V 0 = E (x0 0)
or V = Ex0
Note the negative sign. As one moves along the direction of electric field, the potential falls.
Hence, optionBis correct.
2. Given, electric field, E = 13 N/C
Electric potential, V = 8J/C
​From formula :
E = where r is the distance of the point from charge
13 =
V =
8 =
Dividing eq. 1 with eq. 2, we get :
Putting value of r and k = 9 X 109 Nm2C-2 in eq. 2, we get :
8 =
q = 0.54 X 10-9 C