Question

What are the sign and magnitude of a point charge that produces an electric potential of −2.36 V at a distance of 2.93 mm?

Answer 1

Answer:

q = -7.691 × $10^{-13}$ C

so magnitude of charge is 7.691 × $10^{-13}$ C

and negative sign mean charge is negative potential

Explanation:

given data

electric potential = −2.36 V

distance = 2.93 mm = 2.93 × $10^{-3}$ m

to find out

What are the sign and magnitude of a point charge

solution

we know here that electric potential due to charge is

V = $k\frac{q}{r}$       ..............................1

here k is coulomb force that is 8.99 ×$10^{9}$ Nm²/C² and r is distance and q is charge  and V is electric potential

put here all value we get q  in equation 1

V = $k\frac{q}{r}$

-2.36 = $8.99*10^{9} \frac{q}{2.93*10^{-3}}$

q = -7.691 × $10^{-13}$ C

so magnitude of charge is 7.691 × $10^{-13}$ C

and negative sign mean charge is negative potential