On the x-y plane, two equal point charges q are located at the points (a, 0) and

Question

On the x-y plane, two equal point charges q are located at the points (a, 0) and (-a, 0). Find the magnitude of the electric field at the point (0.y). Find the value of y where the electric field is maximum. Sketch the electric field as a function of y. An electric dipole consists of a point charge -q at the point (0, a) and a point charge q at the point (0, -a). Find the electric field at the point (x, 0). Three point charges q, q. and -2q occupy three comers of a square of side a as shown. Find the magnitude and direction of the electric field at (a) the unoccupied comer, and (b) the center. Three point charges q, q, and -2q occupy the vertices of an equilateral triangle of side a. Find the electric field at the center of the triangle. A water droplet of radius 0.018mm remains stationary in air. If the downward electric field is 300N/C, how many excess electrons must the droplet have?

Explanation / Answer

Here, the electrical force due to electric field = E*q, where q = number of electrons*1.60x10^-19C

And the weight = density time volume*g
= 1000kg/m^3*(4/3*r (in meters)^3)*g

Now, the droplet remains stationary in air.

Therefore, force due to electric field = weight of the droplet

So, n *1.60x10^-19C *300 = 1000kg/m^3*(4/3*r (in meters)^3)*g

=> n = ( 1000kg/m^3*(4/3*(0.000018)^3)*g)/(300*1.60x10^-19) = 5.09 x 10^5

So, no. of excess electrones = 5.09 x 10^5

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