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taurus [48]
3 years ago
12

I need help on 7 &8

Physics
1 answer:
Anettt [7]3 years ago
6 0
I believe it’s a then d
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Two point charges, 3.4 μC and -2.0 μC , are placed 5.0 cm apart on the x axis. Assume that the negative charge is at the origin,
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The electric field is zero at x = -16.45cm

Data;

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<h3>The Electric Field at point 0</h3>

As the 3μC is larger than -2.0μC  and the charges are opposite sign. The electric field will be zero at the negative axis.

Let the point be at x.

For an electric field to be equal to zero;

k(\frac{q_1}{d_1})^2 + k(\frac{q_2}{d_2})^2 = 0\\\frac{3.4}{(5-x)^2} - \frac{2}{x^2} = 0\\

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\frac{3.4x^2 - 2(5-x)^2}{(5-x)^2x^2}= 0\\ 3.4x^2 - 2(5-x^2) = 0\\3.4x^2 - 50 - 2x^2 + 20x = 0\\1.4x^2 +20x - 50 = 0

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x = -16.45cm

The electric field is zero at x = -16.45cm

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