A-5.0 mC charge travels due south under the influence of a 10 N force when place

Question

A-5.0 mC charge travels due south under the influence of a 10 N force when placed into a electric field. What force would a 10 mc charge experience if placed in the same field and in which direction would it move? what is the magnitude of a point charge that would create an electric field of 1.00NIC at points 1.00 m away? Consider a fixed point charge of +2.00 HC. What is the magnitude and direction of the electric field at a point P, a distance of 0.100 m away? An electric field has an electric field strength 6000. N/C at a distance of 1.5 m. What is the strength of the field at a distance of 6.0 m?

Explanation / Answer

formula for force acting on charge is

F = Eq

F1 = Eq1

F2 = E q2

F2/F1 = E q2/E q1

F2 = F1 ( q2/q1)

= 10 ( 10/5)

= 20 N

charge along east ward direction

(5)

E = kq/r^2

q = Er^2/k

=1(1)^2/9 * 10^9

=0.11 * 10^-9 C

(6)

E = 9* 10^9 ( 2 * 10^-6)/(0.1)^2

=1.8 N/C

(7)

E1 = kq/r1^2

E2 = kq/r2^2

E2/E1 = (r1/r2)^2

E2 = 6000( 1.5/6)^2

=375 N/C

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