JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 1)
There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at $$P,$$ in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60o with the direction of the field ?
589.5 V
589.2 V
589.4 V
589.6 V
Forklaring
Potential gradient,
$$\Delta $$V = E. d
$$ \Rightarrow $$$$\,\,\,$$ 589.8 $$-$$ 589.0 = (E d)max
$$ \Rightarrow $$$$\,\,\,$$ (E d)max = 0.8
$$\therefore\,\,\,$$ $$\Delta $$V = E d cos$$\theta $$
= 0.8 $$ \times $$ cos60o
= 0.4
$$\therefore\,\,\,$$ Maximum potential on the sphere = 589.4 V
$$\Delta $$V = E. d
$$ \Rightarrow $$$$\,\,\,$$ 589.8 $$-$$ 589.0 = (E d)max
$$ \Rightarrow $$$$\,\,\,$$ (E d)max = 0.8
$$\therefore\,\,\,$$ $$\Delta $$V = E d cos$$\theta $$
= 0.8 $$ \times $$ cos60o
= 0.4
$$\therefore\,\,\,$$ Maximum potential on the sphere = 589.4 V