Question

1.) What is the strength and direction of the electric field 3.74 cm on
the left hand side of a 9.1 nC negative charge?

[Electrostatic constant, k = 9.0 x 109 N • m2 / C2]

Answer 1

The strength of the electric field produced by a point charge is given by:
$E(r) = k \frac{q}{r^2}$
where
k is the electrostatic constant
q is the charge
r is the distance from the charge at which the field is calculated

Using $q=9.1 nC=9.1 \cdot 10^{-9} C$ and $r=3.74 cm=0.0374 m$, we find
$E=k \frac{q}{r^2}= (9\cdot 10^9 Nm^2C^{-2})\frac{9.1 \cdot 10^{-9}C}{(0.0374 m)^2} =5.86 \cdot 10^4 N/C$

Concerning the direction of the field: the electric field produced by a point charge is radial, and for a negative charge, the field lines are directed toward the source of the field, so toward the negative charge. Since we are on the left side of the charge, the electric field direction is toward the right.