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Ilia_Sergeevich [38]
3 years ago
14

Two positive point charges with charges q1 and q2 are separate by a distance of 30cm. A third positive charge qo is placed betwe

en 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.
Physics
1 answer:
vivado [14]3 years ago
5 0

Answer:

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

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

\dfrac{kq_1q_0}{0.08^2}=25\\\\q_1=59.25\times10^{-7}\ \rm C

similarly

\dfrac{kq_2q_0}{0.22^2}=10\\\\q_2=179.25\times10^{-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\\=\dfrac{kq_1}{0.08^2}-\dfrac{kq_2}{0.22^2}\\\\=\dfrac{9\times10^9\times59.25\times10^{-7}}{0.08^2}-\dfrac{9\times 10^9\times 179.25\times 10^{-7}}{0.22^2}\\\\=5\times 10^6\rm N/C

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

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