Physics, asked by Narendrabora85, 9 months ago

please provide solution
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Answered by desichorapriyanshu
1

hey mate here ur answer :-

E = kQ/r², r = 3.85 m since that is the mid-point

E = kQ/r², r = 3.85 m since that is the mid-pointE(net) = (9e9)(12e-6)/(3.85)² + (9e9)(15e-6)/(3.85)²

E = kQ/r², r = 3.85 m since that is the mid-pointE(net) = (9e9)(12e-6)/(3.85)² + (9e9)(15e-6)/(3.85)²= 16,394 N/C towards from the -12 µC charge

E = kQ/r², r = 3.85 m since that is the mid-pointE(net) = (9e9)(12e-6)/(3.85)² + (9e9)(15e-6)/(3.85)²= 16,394 N/C towards from the -12 µC chargeThe 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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