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