Three point charges are arranged along the x axis. Charge q1=-4.50 nC is located at x=0.

step 1 The given problem here this is the x -axis. This is the origin at which we are placing a charge

Three point charges are arranged along the x axis. Charge...

Three point charges are arranged along the $x$ axis. Charge $q_{1}=-4.50 \mathrm{nC} \quad$ is $\quad$ located $\quad$ at $\quad x=0.200 \mathrm{~m}, \quad$ and $\quad$ charge $q_{2}=+2.50 \mathrm{nC}$ is at $x=-0.300 \mathrm{~m}$. A positive point charge $q_{3}$ is located at the origin. (a) What must the value of $q_{3}$ be for the net force on this point charge to have magnitude $4.00 \mu \mathrm{N}$ ? (b) What is the direction of the net force on $q_{3} ?$ (c) Where along the $x$ axis can $q_{3}$ be placed and the net force on it be zero, other than the trivial answers of $x=+\infty$ and $x=-\infty ?$

Key Concepts

Electrostatic Equilibrium

Electrostatic equilibrium refers to a condition where the net force acting on a charge is zero, meaning that there is no resultant acceleration. In problems involving multiple charges, finding positions where the electric forces cancel each other is a common application of the equilibrium concept, providing insights into stable and unstable configurations.

Vector Addition of Forces

Since electric forces are vector quantities, determining the net force on a charge involves adding the forces from individual charges both in magnitude and direction. This requires careful consideration of the sign and direction of each contributing force, ensuring proper resolution of components along the relevant axes.

Coulomb's Law

Coulomb's Law quantifies the electric force between two point charges, stating that the force is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This law is foundational in electrostatics, allowing the calculation of forces between any two charged objects.

Superposition Principle

The superposition principle in electrostatics holds that the net force on any charge due to a collection of other charges is the vector sum of the individual forces exerted by each charge independently. This principle is essential for analyzing systems with multiple charges, as it allows the cumulative effect of fields or forces to be determined.

Transcript

00:02 The given problem here this is the x -axis.

00:09 This is the origin at which we are placing a charge q3.

00:19 A charge q1 has been kept here which is having a negative charge.

00:30 Let this point is a along the x -axis at which this charge has been kept minus q1.

00:38 The distance of this point a from the origin is 0 .200 meter.

00:47 Then another charge, positive in nature, plus q2 has been kept here.

00:56 Suppose at a point b, whose distance from the origin is minus 0 .300 meter.

01:05 Now it is given that the next.

01:08 Net force acting at this charge q3, q2 these two charges, charge q2 put at b and charge minus q1 put at a is given as that would be f at o and that is equal to 4 .00 micronuton.

01:27 The value of q1 is minus 4 .50 nanoculum.

01:40 And value of q2 is plus 2 .50 nanoculum.

01:49 So, using this expression f at o, there are two forces at this charge q3.

01:58 As q3 is a positive charge, it will be attracted by this minus q1 by a force, f at o due to a.

02:12 And it will be repelled by this.

02:14 Positive q2 f at o due to b both are in the same direction hence net force f o will be given by the sum of f at o due to a plus f at o due to b so using coulalams law for the electrostatic force for f o a this is k into q1 into q 3 divided by oa, which is 0 .20 meter to the whole square.

02:56 Plus for f -o -b, a again, that is q2 into q3.

03:03 At o, this is 3.

03:05 At b, this is q2.

03:09 This stands 0 .3 -0 to the whole square.

03:12 And this force will come out to be 4 .00 micronuton or 4 .00 into 10 dash bar minus 6 newton.

03:22 Now taking the common things out which include the value of k which is 9 into 10 dash bar 9 then q3 that can also be taken out as a common living inside for q1 this is 4 .50 only the magnitude as the direction already we have used in the diagram 4 .50 into 10 dash 4 minus 9 kulam divided by 0 .04 meter making a square of 0 .2.

04:01 And here this is 2 .50 into 10 dash per minus 9 divided by the square of 0 .3 which comes out to be 0 .09.

04:17 And it should be 4 into 10 dash par minus 6...

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