Question

A particle has charge -1.95 nC. (a) Find the magnitude and direction of the electric field due to this particle at a point 0.225 m directly above it magnitude direction | Select ' N/C (b) At what distance from this particle does its electric field have a magnitude of 10.5 N/C?

Answer 1

Answer:

a)

346.67 N/C, downward

b)

1.3 m

Explanation:

(a)

q = magnitude of charge on the particle = 1.95 x 10⁻⁹ C

r = distance of location from the charged particle = 0.225 m

E = magnitude of electric field at the location

Magnitude of electric field at the location is given as

$E = \frac{kq}{r^{2}}$

Inserting the values

$E = \frac{(9\times 10^{9})(1.95 \times 10^{-9})}{(0.225)^{2}}$

E = 346.67 N/C

a negative charge produce electric field towards itself.

Direction :  downward

(b)

E = magnitude of electric field at the location = 10.5 N/C

r = distance of location from the charged particle = ?

q = magnitude of charge on the particle = 1.95 x 10⁻⁹ C

Magnitude of electric field at the location is given as

$E = \frac{kq}{r^{2}}$

Inserting the values

$10.5 = \frac{(9\times 10^{9})(1.95 \times 10^{-9})}{r^{2}}$

r = 1.3 m