# Three identical point charges of charge q = 5.6 \muC are placed at the vertices (corners) of an...

## Question:

Three identical point charges of charge q = 5.6 {eq}\mu {/eq}C are placed at the vertices (corners) of an equilateral triangle. If the side of triangle is a = 5.1m, what is the magnitude, in N/C, of the electric field at the point P in one of the sides of the triangle midway between two of the charges?

## Electric Field:

Electric field is defined as the force per unit positive charge that will act on this unit charge at a point of observation. The direction of electric field is away from the positive and towards negative electric charge. The electric field due to a system of charges is the vector sum of all individual electric fields produced by individual charges at the point of observation.

## Answer and Explanation:

The electric field at a point midway between two charges is determined by only the third charge, since the fields due to two other charges completely cancel each other. The strength of the field is:

{eq}E = \dfrac {kq}{d^2} {/eq}

Here

- {eq}k = 9\times 10^9 \ N/(m^2\cdot C^2) {/eq} is the Coulomb's constant;

- {eq}q = 5.6\times 10^{-6} \ C {/eq} is the charge

- {eq}d = \dfrac {a\sqrt{3}}{2} {/eq} is the distance to the point from the charge at the opposite corner;

- {eq}a = 5.1 \ m {/eq} is the side of the triangle;

Computing, we obtain:

{eq}E = \dfrac {4\cdot 9\times 10^9 \ N/(m^2\cdot C^2) \cdot 5.6\times 10^{-6} \ C}{3 \cdot (5.1 \ m)^2} \approx 2.58\times 10^{-3} \ N/C {/eq}

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