91Ó°ÊÓ

The potential difference in uniform electric field, $∆V_1=74V$from the origin to (0,0,10) m.

The spherical shell uniform radius, r = 5 m

The spherical shell charge, Q = -3530 nC

The expression to calculate the potential difference in terms of electric field is given as follows.

$â–³V=Ed$.......................(i)

Here, is the electric field and is the distance between the two points.

The expression to calculate the potential difference is given as follows.

$â–³V=\frac{Q}{4{\mathrm{ππÎ\mu }}_0}(\frac{1}{r_1}-\frac{1}{r_2})$ .............................(ii)

Consider the system with uniform electric field and spherical shell.

The potential difference from the origin to (0,0,5) m is sum of the potential difference due to uniform electric field and due to shell.

$â–³V=â–³V_2+â–³V_{sphere}$ ..................................(iii)

Since the potential difference is directly proportional to the distance between the two points. Because the distance between the two point is reduced by halved then the potential difference due to uniform electric field is given by,

$â–³V_2=\frac{1}{2}â–³V_1$

Substitute 74 V for $â–³V_1$into above equation.

$\begin{array}{l}△V_2=\frac{1}{2}×74\\ △V_2=37V\end{array}$

The total charge is concentrated on the surface of the spherical shell. The potential inside the shell is constant at each point. Therefore, the potential between origin and (0,0,5) m is zero.

$V_{sphere}=0\mathrm{V}$

Calculate the potential difference.

Substitute 0 V for $â–³V_{sphere}$and 37 V for $â–³V_2$into equation (iii).

$\begin{array}{ll}â–³V=37+0& \\ â–³V=37V& \end{array}$

Hence the potential difference from origin to (0,0,5) m is 37 V.

Calculate the potential difference due to uniform electric field.

$â–³V_{field}=â–³V_1+â–³V_2$

Substitute 74 V for $â–³V_1$and 37 V for $â–³V_2$into above equation.

$$\begin{gathered} â–³V_{Field}=74-37 \\ â–³V_{Field}=37V \end{gathered}$$

Calculate the potential difference due to spherical shell.

Substitute $9×{10}^{9}N.m^2{C}^{-2}$for $\frac{1}{4{\mathrm{Ï€Î\mu }}_0}$ and -3530 nC for Q, 10 m for r₁and 5 m for r₂ into equation (ii).

$$\begin{gathered} △V_{sphere}=9×{10}^9\left(\frac{1}{10}-\frac{1}{5}\right) \\ △V_{sphere}=9×{10}^9×\left(-3530×{10}^{-9}\right)\left(-0.1\right) \\ △V_{sphere}=3177V \end{gathered}$$

The potential difference from (0,0,5) m to (0,0,10) m is the sum of the potential difference due to uniform field and spherical shell.

$â–³V=â–³V_{Field}+â–³V_{Sphere}$

Substitute 3177 V for $â–³V_{Sphere}$and 37 V for $â–³V_{Field}$into above equation.

$\begin{array}{l}â–³V=37+3177\\ â–³V=3214V\end{array}$

Hence the potential difference from the position of (0,0,5) m to a position (0,0,10) m is 3214 V.