You previously calculated the electric force on a charge...

You previously calculated the electric force on a charge placed at the center of the circle, now calculate the net electric field if the charge is removed. For the 3 charges on a circle of radius r = 2.00 cm, where q = 6.00 nC.

Transcript

00:01 In this question we are using a figure the total charge on the ring the capital q is given us plus 28 nanocolon that is total charge is q we are also given a point charge in the center q is minus 22 nanocolon at the center we are given a the distance from the center to the point your electric field is to be find this 33 meter let this will be x and we are given the radius of the ring a is 0 .3 meter so this will be a you have to find out the electric field at a point p here so let's take the angle this be theta then this will be also theta if this is also theta then this will be theta we can solve the components of electric field so because of the queue here when a test charge is placed there we get if electric field de in this direction and de in this direction so if this is theta then one of the component of de will be in this direction and other will be in this direction if this is angle theta then one of the component will be in this direction so this will be de sine theta component this will be d .e.

01:50 Sine theta component and both of this is d .e.

01:56 Cos theta component as q is the charge.

02:01 Now we can write the angle cos theta will be able to adjacent by hypoponesis root of x square plus a square so the distance this will be root of x square plus a square now now we can write the electric field so the electric field point p is d e cos theta when we are taking the element then we can write the expression for the d e cost theta will be equal to k into the element charged dq divided by the distance from the charge to the point is root of x square plus a square whole square into cos theta is x by root of x square plus a square rewriting the time we get k into x dq divided by x square plus a square all raised to three by two now we can find out the electric field total left -tube will be equal to integral b e cos theta that is e will be equal to integral k into x dq divided by x square plus a square whole raise to 3 by 2 performing integration we get e will be equal to k x into q divided by x square plus a square all reach to 3 by 2...

Question

You previously calculated the electric force on a charge placed at the center of the circle, now calculate the net electric field if the charge is removed. For the 3 charges on a circle of radius r = 2.00 cm, where q = 6.00 nC.

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