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? A. 25.2×106 N/C directed towards the positive charge B. 25.2×106 N/C directed towards the negative charge C. 25.2×104 N/C directed towards the negative charge D. 25.2×105 N/C directed towards the negative charge E. 25.2×105 N/C directed towards the positive charge The correct answer is B, but I was hoping someone could work it out and explain why!

Explanation / Answer

GIVEN: q1=20*10^-6C,q2=-8*10^-6C,distance=20cm=>half way=d=10cm=0.1m.

E=E1+E2
E1=kq1/d^2
E2=kq2/d^2
then
E=(k/d^2)(q1+|q2|)
E=(9x10^9/10^-2)(20+8)x10^-6
E=25.2x10^6 N/C directed towards the negative charge

Answer is B

The reason we add the e-fields together is because one charge is negative and the other charge is positive, so the two charges are attractive. At the mid-point, the e-fields of the two charges are additive and points towards the negative charge.

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