Question

Find the electric field (magnitude and direction) for the following situations:

A) At a point 1.0x10-3 m away from a 5.00 pC charge. (4.5x104 N/C, away)

B) Halfway between a 6.0 ?C and a -8.0 ?C charge, which are 10.0 m apart. (5040 N/C towards -8 ?C*)

C) 1.0 m from the end of a 3.5 m long uniformly charged rod, which has a 4.0 nC net charge. (8.00 N/C away from rod)

Answer 1

A) Electric field (magnitude and direction) at a point 1.0x10-3 m away from a 5.00 pC charge

q = 5 * 10^-12 C

r = 1 * 10^-3 m

k = 9 * 10⁹ N.m² / C² is the Coulomb's force constant.

Magnitude of the electric field is given by,

E = k q / r²

E = 9 * 10⁹ * 5 * 10^-12 / (1 * 10^-3)²

E = 4.5 * 10⁴ N/C and this field directs away from the point at which we are calculating the electric field. This is because the charge is positive and electric field due to positive charge points outward, that is, it is directed away from the charge.

B) Electric field (magnitude and direction) halfway between a 6.0 C and a -8.0 C charge, which are 10.0 m apart.

q₁ = 6.0 C = 6 * 10^-6 C

q₂ = -8.0 C = -8 * 10^-6 C

Separation between the charges is d = 10 m

Thus, distance of each charge from the point at which we are measuring electric field is, r = 5 m

So, magnitude of the electric field at a point 5 m from q₁ and q₂ is,

E = kq₁/ r² + kq₂/ r²

E = 9 * 10⁹ * [6 * 10^-6 + 8 * 10^-6] / (5)²

E = 5.04 * 10³ N/C = 5040 N/C

Direction:

Electric field due to +6.0 C is directed towards the point at which we are measuring the electric field and electric field due to -8.0 C is directed away from the point at which we are measuring the electric field and towards -8.0 C. This effectively means, the net electric field is directed towards -8.0 C.

C) 1.0 m from the end of a 3.5 m long uniformly charged rod, which has a 4.0 nC net charge

Length of the rod, L = 3.5 m

distance of the point at which we are measuring the field from one end of the rod is, a = 1 m

Total charge on the rod, Q = 4 * 10^-9 C

Electric field due to an uniformly charged rod at a point along it's axis is given by,

E = kQ/L * [1/a - 1/(L+a)]

E = 9 * 10⁹ * 4 * 10^-9 [1/1 - 1/ (3.5+1)] / 3.5

E = 8 N/C

Direction:

Since the rod has total positive charge, the field should be pointing away from the rod.