A +2.00 nC point charge is at the origin, and a second -5.00 nC point charge is on the x -axis a

step 1 In the given problem here this is the x -axis. This is the origin x -axis and y -axis. One of th

A +2.00 nC point charge is at the origin, and a second -5.00...

Key Concepts

Coulomb's Law

Coulomb's Law quantifies the electric force between two point charges. It states that the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This law is fundamental in calculating both the magnitude and the direction of forces in electrostatic problems.

Electric Field Due to a Point Charge

The electric field due to a point charge is a measure of the force per unit charge that a test charge would experience at a given point in space. The field is radially directed away from a positive charge and toward a negative charge, and its magnitude is determined by Coulomb’s law by dividing the force by the test charge.

Principle of Superposition

The principle of superposition states that when multiple charges are present, the total electric field at any point is the vector sum of the electric fields created by each individual charge. This concept is essential in solving problems where more than one charge contributes to the net field.

Electric Force on a Charge

The electric force experienced by a charge in an electric field is obtained by multiplying the magnitude of the charge by the electric field at that point. This relationship, expressed as F = qE, is crucial for understanding how charges interact with electric fields and for calculating the forces exerted on them.

Transcript

00:01 In the given problem here this is the x -axis.

00:07 This is the origin x -axis and y -axis.

00:20 One of the charged particle has been kept here at the origin which is having a value of plus 2 .00 nanoculum and another chart particle has been kept at a position x -culem.

00:42 Is equal to 0 .80 meter.

00:46 So let it be this.

00:50 The another chart particle.

00:52 And suppose its position is a.

00:56 So we can say this oa is 0 .80 meter and the value of this charge has been given as minus 5 .00 nanoculum.

01:15 We have to find electric field with the magnitude and direction at few given points in the first part of the problem and the first point at which we have to find electric field is b which is at a distance of here this is point b which is at a distance of 0 .2 0 0 .0 meter from the origin.

01:50 Another point is c, which is at a distance of, here this is, point c, which is at a distance of 1 .2 meter again from the origin.

02:09 And the third point is here, d, which is at a distance of 0 .200 meter, at negative x -axis because the position of d has been given as minus 0 .200 meter.

02:29 It means this is at negative x -axis.

02:33 So in the first sub -part of first part of the problem, we have to find electric field at point b.

02:45 And at this point b, electric fields, there are two electric fields, one because of positive.

02:52 2 .00 nanoculum charge and electric field due to positive charge is always away.

03:01 So here this is electric field and electric field due to the negative charge is towards the negative charge.

03:14 So there are two electric fields at b.

03:16 One is e at b due to o which is away from the positive charge and another is e at b due to a.

03:27 Which is towards the negative charge.

03:30 So the net electric field is eb is equal to e .b.

03:39 Plus e.

03:40 B .a.

03:41 Now we will find the values of these two electric fields separately.

03:46 So e.

03:47 Bo will be given as k into charge which is 0, which is 2 .00 into charge, which is 2 .00 into 10 dash par minus 9 kulum for nano the conversion factor is 10 dash per minus 9 divided by the distance ob whole square so for k the value of k is 9 into 10 dash of 9 into 2 into 10 dash bar 9 divided by ob square which is 0 .2 whole square so here it comes out to to be 9 into 2 because 10 dash par plus 9 and minus 9 will be cancelled out.

04:38 So this is 9 into 2 in the numerator divided by 4 into 10 dash par minus 2 newton per coulum or it comes out to be 4 .5 into 10 dash par 2 newton per coulan then electric field e at b due to a and that is having the same formula, k, but the charge is 5, only the magnitude of the charge, 5 into 10 dash power minus 9 column divided by the distance means ab to the whole square distance between a and b.

05:19 So for k, this is 9 into 10 dash to 9 into 5 into 10 dash to the power minus 9 divided by the square of ab, which is oa minus ob to the whole square.

05:35 In the numerator again, it becomes 45 divided by for oa, this is 0 .8.

05:43 And for ob, this is just 0 .2 to the whole square.

05:48 So finally, it becomes e at b due to a is equal to 45 divided by the square.

05:58 Of 0 .6 which is 36 into 10 dash bar minus 2.

06:05 So finally this electric field comes out to be 1 .25 into 10 dash per 2 newton per coulame hence the answer for this first part e b is equal to e b o plus e b a means this is 4 .5 plus 1 .25 into 10 dash per 2 .2.

06:29 Newton per kulam or we can say this is e .b is equal to 5 .75 into 10 dashed part 2 newton per kulam answer for the first sub part of the first part of the problem now.

06:52 In the second subpart we have to find electric field at the point c and if we look at this point c again here at this point there will be two electric fields one because of negative charge which is towards the negative charge and this is e at c because of a and another is due to this positive charge which is away and this is e at c because of o as it is clear from the figure this e at c because of a should be greater as the charge is closed while this e because of e at c because of o should be smaller value because the distance is very large.

07:41 Hence the net electric field at c will be equal to difference of the two electric fields so it will be given as e...

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