4. In the figure below, if Q = +3.20 × 10^-19 C, q = 1.60 ×...

4. In the figure below, if Q = +3.20 × 10$^{-19}$ C, q = 1.60 × 10$^{-19}$ C, and a = 2.00 cm, what is the electrostatic force on the particle at the origin (+Q) due to the other charged particles? You may write your answer in unit vector notation.

Transcript

00:02 Hello, in the question we have given this configuration of charges.

00:06 Now what we have to do is we have to find out the magnitude of the electrostatic force on this charge small q.

00:16 So the values of this capital q is given which is 41 microcoulomb, then small q is 7 .3 microcoulomb and the distance d is 25 centimetres.

00:26 So we have to find out what this force on this charge minus this small q.

00:35 So let us first have an revision of what will be the forces, how the forces will act.

00:40 Now due to this capital q charge, the force will be repulsive.

00:44 So considering that this is positive, this is positive, so the charge, this will be the force direction.

00:51 So this force, let us call it as fq, the force due to the charge q.

00:56 Then force due to this charge 2q will be in this direction.

01:01 Again repulsive, so this let us name it as f2q.

01:05 So this fq, fq will be equal to kqq divided by d square, because the distance between the charge capital q and small q is d.

01:22 So now let us plug the value.

01:25 So k is 9 into 10 raised to 9 into q, capital q is how much? 41 into 10 raised to minus 6 coulombs.

01:40 It is given in micro, so we are converting into coulombs into 7 .3 into 10 raised to minus 6 divided by, sorry, divided by d square.

01:57 D is 25 centimetres, so we have to convert it into metres, so it will be 0 .25 square.

02:04 And if we calculate this, so fq will turn out to be 43099 .2 into 10 raised to minus 3 newtons.

02:20 I haven't used this 10 raised to minus 3 out here.

02:24 So, or we can write fq as 43 .099 newtons.

02:35 So this is the force due to the charge capital q.

02:38 Now force due to this 2q charge will be how much? k into 2q times q divided by the distance is 2d and square of this.

02:52 So if we calculate this, k is 9 into 10 raised to 9 into 2 into q is 41 into 10 raised to minus 6 into 7 .3 into 10 raised to minus 6 divided by 2 into 0 .25 square.

03:17 And if we calculate this, so this force will turn out to be how much? so force due to this 2q charge will be also, one more thing is left over here, direction.

03:31 See, fq is in the positive x direction.

03:34 It is going in this way...

Question

4. In the figure below, if Q = +3.20 × 10$^{-19}$ C, q = 1.60 × 10$^{-19}$ C, and a = 2.00 cm, what is the electrostatic force on the particle at the origin (+Q) due to the other charged particles? You may write your answer in unit vector notation.

Please give Ace some feedback