An electric dipole consists of a positive and negative charge. The two charges a

Question

An electric dipole consists of a positive and negative charge. The two charges are aligned such that a horizontal line passes through each of the charges as shown in the diagram and the distance separating the two charges is d = 46.50 cm. The positive charge has a charge of +18.80 ?C and the charge on the negative charge is -18.80 ?C. Calculate the magnitude of the electric field at a point a distance d = 46.50 cm directly to the left of the negative charge.

I am stumped here. First I thought you could just use the electric field equation: k |q|/ r^2, find the vecors of the positive and negative charge respectively, and then add them, but I think I am way off.

Explanation / Answer

Since only the magnitude of the electric field has been asked int he question... find the electric field of each charge seperately and subtract the smaller one from the bigger.

E.F due to negative charge = (K x 18.8 x (10^-6) )/ 46.5^2

E.F due to positive charge = (K x 18.8 x (10^-6))/ 92^2

difference of the charges will be = (K x 18.8 x (10^-6)) x 3.443 x 10^-4

that will come out to be = 58.255 N/C

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