Question

Two positive point charges with charges q1 and q2 are separate by a distance of 30cm. A third positive charge qo is placed between the two charges at 8cm on the right of the charge q1. See picture. The charge qo is equal to 3 µC (micro-coulombs = 10-6 C). The Coulomb force exerted by q1 on qo is 25 N and the Coulomb force exerted by q2 on qo is 10 N. Determine the net electric field in magnitude and direction on the charge qo due to the other two charges.

Answer 1

The magnitude of net electric is
`5* 10^6\ \rm N/C` from charge
`q_1` to
`q_2` along the line joining them.

Step-by-step explanation:

Given:

Distance between the charges =30 cm

Magnitude of charge
`q_0 =3\ \rm \mu C`

Force exerted by
`q_1` on
`q_0`=25 N

Force exerted by
`q_2` on
`q_0`=10 N

Now according to coulombs Law we have

`(kq_1q_0)/(0.08^2)=25\\\\q_1=59.25*10^(-7)\ \rm C`

similarly

`(kq_2q_0)/(0.22^2)=10\\\\q_2=179.25*10^(-7)\ \rm C`

Noe the electric field at eh position of charge
`q_0` is given by

Let
`E_1` be the electric field due to charge
`q_1`` and
`E_1` bet he electric field due to charge
`q_2``

then The net electric Field at the point is given by

`E_(net)=E_1-E_2\\=(kq_1)/(0.08^2)-(kq_2)/(0.22^2)\\\\=(9*10^9*59.25*10^(-7))/(0.08^2)-(9* 10^9* 179.25* 10^(-7))/(0.22^2)\\\\=5* 10^6\rm N/C`

The direction of electric Field is from charge
`q_1` to
`q_2` along the line joining them.