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Ball A has a charge of 1 C, while ball B has a charge of -2 C. The balls are placed 2 m apart, and held in place. What is the net electric field from the two balls at a point half-way between them? Remember that k-8.988x10A9 N mA2/CA2. Hint: Electric fields point toward negative charges and away from positive charges. Be very careful with your directions. 9.0x10A9 Towards B O a сь Ос. 9.0x10A9 Towards A b. 2.7x10 10 N/C Towards B 2.7x10A10 N/C Towards A d.

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Answer #1

Field of ball A being positively charged , is away from it , that towards B. Field of ball B, being negatively charged, is towards itself that is towards B. As field due to both balls is towards B, net field is sum of magnitude of two fields and towards B.

Field of Ball A = kqa /1^2 = 8.988×10^9

Field of ball B = k qb /1^2 = 8.988×10^9*2

Net field is sum of the two = 2.7×10^10

Option C is correct.

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