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If a negative test charge of magnitude 2 x 10-6 is placed at this point what is the force experienced by the charge. |
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Answer» EA = \(\frac1{4\pi\varepsilon_0}\frac{q}{r^2}\) EA = 9 x 109 x \(\frac{2}{5^2}\) EA = \(\frac{18\times10^9}{25}\) ⇒ 0.72 x 109 N/C EB = \(\frac1{4\pi\varepsilon_0}\frac{q}{r^2}\) = 9 x109 x \(\frac2{5^2}\) EB = \(\frac{18}{25}\times10^9\) EB = 0.72 x 109 N/C E = EA + EB E = 0.72 x 109 + 0.72 x 109 E = 1.44 x 109 N/C F = qE F = 6 x 1.44 x 109 F = 8.64 x 109 N |
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