What is the magnitude of the electric field 21.3 cm directly above an isolated 2.93×10−5 C...

Question

Question

What is the magnitude of the electric field 21.3 cm directly above an isolated 2.93×10⁻⁵ C charge?

What is the direction of the electric field?

science physics

Answer 1

Answer #1

The electric field for a charge q, at a distance of r from it is given by the relation: Kq/ r^2

Futher, the electric field lines for a negative charge converge towards the charge while for a positive charge, they diverge outwards.

Here, q = 2.93 x 10^-5 C

and r = 21.3 cm

That is, E = 8.99 x 10^9 x 2.93 x 10^-5 / 21.3 * 21.3 * 10^-4 = 0.0581 x 10^8 N/C

Answer 2

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