Question

Three charges 10 microcoulomb ,5 microcoulomb and -5 microcoulomb are placed in air at three corners A,B,C of an equilateral triangle of side 100cm .find the resultant force experienced by charge placed at A.

Answer 1

hope it will help you.

Answer 2

Answer:

The resultant force experienced at point A is found to be 0.45N.

Explanation:

Here, we are given that there are three charges of magnitude: 10μC, +5μC, and -5μC at the points A, B, and C of an equilateral triangle of side 100cm, i.e., 1m.

Now, at the point A, the charge placed is: 10μC

As the surrounding two charges have same magnitude but different signs, we can use the following expression to calculate the resultant force on point A:

$F_{net}=\frac{kq_1q_2}{r^2}$

where k is a constant having a value of $9\times10^9$ in SI units,

and r is the distance between the charges, in this case being the side of the triangle.

Now, substituting the known values, we get the expression as:

$F_{net}=\frac{9\times10^9\times 5\times 10^{-12}\times 10}{1^2}$

$F_{net}={450\times10^{-3}} N$

or we can say that:

$F_{net}=0.45 N$

Thus, the net force on point A is 0.45N.

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