The field E1 at x due to the Ca++ ion acts toward the Cl- ion (or away from the +ve Ca++)

E1 = (kQ)/(k r2) = (9X10-9)(2X1.6X10-19)/(80X2X10-9) = 9.00X108 N/C

The field E2 at x due to the Cl- ion acts toward it (or in the same direction as E1)

E2 = (kQ)/(k r2) = (9X10-9)(1.6X10-19)/(80X2X10-9) = 4.50X108 N/C

The total field is 13.5X108 N/C toward the Cl- ion.

The answer is A

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