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Two point charges lie on the x axis. a charge of 6.5 μc is at the origin, and a charge of -9.2 μc is at x=10.0cm. what is the net electric field at (a) x=−4.0cm and at (b) x=+4.0cm?

Respuesta :

vaduz

the electric field due to a charge 'q' at a distance 'r' is given by

E = q/r

Now, we shall calculate electric field at a point due to each charge individually and then use the principal of super position and add these results to find the net electric field at the point. This can be done for any number of charges.

q1 = 6.5 uC

q2 = -9.2 uC

At x = -4cm

r = 4cm

r2 = 14cm

[tex]E = \frac{1}{4.\pi.e} .[q1/r1^2 + q2/r2^2][/tex]

E = 0.3229 N/C

Similarly at x = -4cm

r1 = 4cm

r2 = 6cm

E = 0.135 N/C

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