Two point charges of +20.0 C and -8.00 C are separated by a distance of 20.0 cm.

Question

Two point charges of +20.0 C and -8.00 C are separated by a distance of 20.0 cm. What is the intensity of electric field E midway between these two charges?

Two point charges of +20.0 C and -8.00 C are separated by a distance of 20.0 cm. What is the intensity of electric field E midway between these two charges?

25.2 × 106 N/C directed towards the positive charge 25.2 × 106 N/C directed towards the negative charge 25.2 × 105 N/C directed towards the negative charge 25.2 × 104 N/C directed towards the negative charge 25.2 × 105 N/C directed towards the positive charge

Explanation / Answer

Given :-

q1 = 20 x 10^-6 c

q2 = - 8 x 10^-6 c

r = 0.20 m

k = 9 x 10^9

we have to find electric field midway between two points

:- r = 0.10 m

The beautiful thing about electric fields is that they add linearly, this is the law of superposition. So all you have to do is calculate the E-field due to each charge, and then add them. Use the formula E=kq/r and then take the sum over both charges. Since this is a vecor field, r must be positive for one and negative for the other. You should find that when the charges are alike, the magnitude of the field is less than when they are different. Just imagine what a small positive test charge would do.

E = kq/r^2

E = 9.0x10^9*(20x10^-6/0.10^2 -8.0x10^-6/0.10^2)

E = 1.080x10^7 N/C

I got this ans.

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