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There is a -5 nC charge present.A. Determine the magnitude and direction of the electric field at a point 7.3 meters from this charge.B. How far from this charge does the electric field have a magnitude of 2 N/C?

User Jerlam
by
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1 Answer

5 votes

Answer:

(A) E = 0.84 N/C

(B) r = 4.74 meters

Step-by-step explanation:

Given that,

Charge,
q=-5\ nC=-5* 10^(-9)\ C

(a) The magnitude of electric field at a point 7.3 meters from this charge is given by :


E=(kq)/(r^2)


E=(9* 10^9* 5* 10^(-9))/((7.3)^2)

E = 0.84 N/C

We know that electric field due to negative charge is inwards. So, the direction of electric field is towards the charge.

(b) Let at a distance of d meters the electric field have a magnitude of 2 N/C. It is given by :


E=(kq)/(r^2)


r=\sqrt{(kq)/(E)}


r=\sqrt{(9* 10^9* 5* 10^(-9))/(2)}

r = 4.74 meters

Hence, this is the required solution.

User Pankaj Negi
by
8.4k points

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