Question

A thin metallic shell of radius 40 cm has a charge of -25 nC on it. At the center of the sphere is a point charge of 35 nC. What is the electric field 50 cm from the center of the shell?
400 N/C outward
560 N/C outward
360 N/C outward
360 N/C inward
560 N/C inward

Answer 1

Answer on Question #84692 Physics / Other

A thin metallic shell of radius $R = 40$ cm has a charge of $Q = -25$ nC on it. At the center of the sphere is a point charge of $q = 35$ nC. What is the electric field $r = 50$ cm from the center of the shell?

Solution:

The Gauss's law for the electric flux through a closed surface

$$\iint \mathbf{E} d\mathbf{A} = \frac{1}{\varepsilon_0} Q_{\mathrm{net}}$$

Let us consider a closed surface as a sphere of radius $r > R$. Thus

$$\iint \mathbf{E} d\mathbf{A} = E \iint dA = E \times 4\pi r^2$$ $$Q_{\mathrm{net}} = Q + q = -25 \, \mathrm{nC} + 35 \, \mathrm{nC} = 10 \, \mathrm{nC}$$

So, electric field

$$\begin{aligned} E &= \frac{Q_{\mathrm{net}}}{4\pi\varepsilon_0 r^2} \\ &= \frac{10 \times 10^{-9} \, \mathrm{C}}{4\pi \times 8.85 \times 10^{-12} \, \frac{\mathrm{F}}{\mathrm{m}} \times 0.5^2 \, \mathrm{m}^2} \\ &= 360 \, \frac{\mathrm{N}}{\mathrm{C}} \end{aligned}$$

**Answer:** $360 \, \frac{\mathrm{N}}{\mathrm{C}}$ outward

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