Question

Two isolated, concentric conducting spherical shells have radii $R_{1} = 0.500 m$ and $R_{2} = 1.00 m,$ uniform charges $q_{1} = + 3.00 μC$ and $q_{2} = + 1.00 μC$ , and negligible thicknesses. What is the magnitude of the electric field $E$ at radial distance $(a)$ $r = 4.00 m$ , $(b)$ $r = 0.700 m$ and $(c)$ $r = 0.200 m$ ? With $V = 0$ at infinity, what is $V$ at $(d)$ $r = 4.00 m$ , $(e)$ $r = 1.00 m$ , $(f)$ $r = 0.700 m$ , $(g)$ $r = 0.500 m$ , $(h)$ $r = 0.200 m$ and $(i)$ $r = 0$ ? $(j)$ Sketch $E (r)$ and $V (r)$ .

Accepted Answer

$(a)$ At $r = 4.00 m$ , electric field (outside both the spheres)

$E = \frac{k q_{1}}{r^{2}} + \frac{k q_{2}}{r^{2}}$

$= \frac{9 \times 10^{9} \times 3 \times 10^{- 6}}{{(4)}^{2}} + \frac{9 \times 10^{9} \times 1 \times 10^{- 6}}{{(4)}^{2}}$

$= 2250 N C^{- 1}$

$(b)$ At $r = 0.700 m$ , (between the two spheres) electric field,

$E = \frac{k q_{1}}{r^{2}} + 0$

$(f)$

$\Rightarrow E = 55102.04 N C^{- 1}$

$(c)$ At $r = 0.200 m$ (inside the smaller sphere)

$E = 0 + 0 = 0$

$(d)$ At $r = 4.00 m$ (outside both the spheres), potential,

$V = \frac{k q_{1}}{r} + \frac{k q_{2}}{r}$

$= \frac{9 \times 10^{9} \times 3 \times 10^{- 6}}{4} + \frac{9 \times 10^{9} \times 1 \times 10^{- 6}}{4}$

$= 9000 V$

$(e)$ At $r = 1.00 m$ (on the surface on second sphere), potential

$r = 0.200 m$

$E (r)$

$= 36000 V$

$(f)$ At $r = 0.700 m$ (between both the spheres), potential

$V = \frac{k q_{1}}{r} + \frac{k q_{2}}{R_{2}}$

$= \frac{9 \times 10^{9} \times 3 \times 10^{- 6}}{0.7} + \frac{9 \times 10^{9} \times 1 \times 10^{- 6}}{1}$

$= 47571 V$

$(g)$ At $r = 0.500 m$ , (on the surface of smaller sphere), potential

$V = \frac{k q_{1}}{R_{1}} + \frac{k q_{2}}{R_{2}}$

$= \frac{9 \times 10^{9} \times 3 \times 10^{- 6}}{0.5} + \frac{9 \times 10^{9} \times 1 \times 10^{- 6}}{1}$

$= 63000 V$

$(h)$ At $r = 0.200 m$ , (inside the smaller sphere)

$V = \frac{k q_{1}}{R_{1}} + \frac{k q_{2}}{R_{2}}$

$= \frac{9 \times 10^{9} \times 3 \times 10^{- 6}}{0.5} + \frac{9 \times 10^{9} \times 1 \times 10^{- 6}}{1}$

$= 63000 V$

$(g)$ At $r = 0$ , (at the common centre)

$V = \frac{k q_{1}}{R_{1}} + \frac{k q_{2}}{R_{2}}$

$= \frac{9 \times 10^{9} \times 3 \times 10^{- 6}}{0.5} + \frac{9 \times 10^{9} \times 1 \times 10^{- 6}}{1}$

$= 63000 V$

$(j)$ We consider electric field along $y$ -axis and distance from common centre $(r)$ along $x$ -axis. Using the calculated values, we draw graph,

We consider electric potential along $y$ -axis and distance from common centre $(r)$ along $x$ -axis. Using the calculated values, we draw graph,

Important Questions on Electric Potential

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