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
.
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 × 10^{10}C, 2m

4.44 × 10^{10}C, 2m

8.89 × 10^{10}C, 0.5m

4.44 × 10^{10}C, 0.5m
QUESTION : 3
Two charges 2×10^{8}C and 3×10^{8}C are located 30 cm apart. Determine the point from the negative charge at which the electric potential is 0.
Options
18 cm

12 cm

16 cm

15 cm
Potential at P
⇒ x= 0.18 m
QUESTION : 4Sixty four water droplets, each of radius 1 mm and charge 10^{10}C combine to form a single large drop. Determine the potential of the bigger drop.
Options
256 V

126 V

280 V

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 E_{0}is directed. Find the potential at x = +x_{0}.
Options

Ex_{0}

Ex_{0}

2Ex_{0}

2Ex_{0}
For a uniform fieldΔV = EΔr. In this case,ΔV = V 0,Δr = x_{0} 0, and E = E_{0}.
Thus,
V 0 = E (x_{0} 0)
or V = Ex_{0}
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
â€‹From formula :
E = $\frac{kq}{{r}^{2}}$ where r is the distance of the point from charge
13 = $\frac{kq}{{r}^{2}}.........\left(1\right)$
V = $\frac{kq}{r}$
8 = $\frac{kq}{r}........\left(2\right)$
Dividing eq. 1 with eq. 2, we get :
$\frac{13}{8}=\frac{1}{r}\phantom{\rule{0ex}{0ex}}r=0.615m$
Putting value of r and k = 9 X 10^{9 }Nm^{2}C^{2} in eq. 2, we get :
8 = $\frac{9X{10}^{9}Xq}{0.615}$
q = 0.54 X 10^{9} C