Question

Two point charges of +60.0 μc and -12.0 μc are separated by a distance of 20.0 cm. what is the magnitude of electric field due to these charges at a point midway between them? (k = 1/4πε 0 = 8.99 × 109 n · m2/c2) two point charges of +60.0 μc and -12.0 μc are separated by a distance of 20.0 cm. what is the magnitude of electric field due to these charges at a point midway between them? (k = 1/4πε 0 = 8.99 × 109 n · m2/c2)

Answer 1

The point midway between the two charges is at

`d= (20~cm)/(2)=10~cm=0.1~m`
from both charges.

1) Let's calculate the electric field generated by the charge
`q_1=60~\mu C=60\cdot10^(-6)~C` at the point halfway between the two charges. It will be

`E_1 = k_e (q_1)/(d^2) = 8.99\cdot 10^9 Nm^2C^(-2) (60\cdot10^(-6)~C)/((0.1~m)^2) = 5.39\cdot10^7~NC^(-1)`

2) Instead, the electric field generated by the charge
`q_2=-12 \mu C =-12\cdot10^(-6)~C` will be

`E_2 = k_e (q_2)/(d^2) = 8.99 \cdot 10^9 Nm^2C^(-2) (-12\cdot 10^(-6)~C)/((0.1~m)^2) =-1.08\cdot10^7~NC^(-1)`

3) The total field is given by the resultant of the two fields. Now, if we assume that the first charge is on the left of the half-way point and the second charge is on its right, the resultant will be the algebric difference of the two fields, therefore the total field is

`E_(tot)=E_1-E_2 = 5.39\cdot10^7~NC^(-1)-(-1.08\cdot 10^7~NC^(-1))=`

`=6.47\cdot10^7~NC^(-1)`
and the verse is towards charge 2 (same verse of
`E_1`).