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What is the magnitude of the electric field 17.1 cm directly above an isolated 1.83Ã10â5 C charge?

User MaM
by
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1 Answer

1 vote

Answer:

Electric field,
E=5.63* 10^(16)\ N/C

Step-by-step explanation:

Given that,

Charge,
q=1.83* 10^5\ C

We need to find the magnitude of electric field 17.1 cm (0.171 m) above an isolated charge. Electric field at a point is given by :


E=(kq)/(r^2)


E=(9* 10^9* 1.83* 10^5\ C)/((0.171\ m)^2)


E=5.63* 10^(16)\ N/C

So, the electric field is
5.63* 10^(16)\ N/C. Hence, this is the required solution.

User Adam Gardner
by
4.3k points