## Related questions with answers

Question

A metal sphere of radius 15 cm has a net charge of $3.0 \times 10^{-8} C$. What is the electric field at the sphere’s surface?

Solutions

VerifiedSolution A

Solution B

Answered 1 year ago

Step 1

1 of 4**Given:**

- Radius: $r = 15 \mathrm{~cm}$;
- Charge: $q = 3 \times 10^{-8} \mathrm{~C}$;
- Potential: $V(\infty) = 0$;
- Potential: $\Delta V = - 500 \mathrm{~V}$;

**Required:**

a) The electric field $E$;

b) The potential $V$;

c) The distance $d$;

Step 1

1 of 4Part a:

$E = k\ \dfrac{q}{r^2} = \dfrac{(9 \cdot 10^9)*(3 \cdot 10^{-8})}{(0.15)^2} = 1.2 \cdot 10^4 \ \mathrm{N/C} = 12 \ \mathrm{kN/C}$

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