Two point charges q1=+2.40 nC and q2=-6.50 nC are 0.100 m apart. Point A is midway between them

Two point charges q1=+2.40 nC and q2=-6.50 nC are 0.100 m...

Two point charges $q_{1}=+2.40 \mathrm{nC}$ and $q_{2}=-6.50 \mathrm{nC}$ are $0.100 \mathrm{~m}$ apart. Point $A$ is midway between them; point $B$ is $0.080 \mathrm{~m}$ from $q_{1}$ and $0.060 \mathrm{~m}$ from $q_{2}$. (See Figure $\left.18.37 .\right)$ Take the electric potential to be zero at infinity. Find (a) the potential at point $A ;$ (b) the potential at point $B ;$ (c) the work done by the electric field on a charge of $2.50 \mathrm{nC}$ that travels from point $B$ to point $A$.

Two point charges $q_{1}=+2.40 \mathrm{nC}$ and $q_{2}=-6.50 \mathrm{nC}$ are $0.100 \mathrm{~m}$ apart. Point $A$ is midway between them; point $B$ is $0.080 \mathrm{~m}$ from $q_{1}$ and $0.060 \mathrm{~m}$
from $q_{2}$. (See Figure $\left.18.37 .\right)$ Take the electric potential to be zero at infinity. Find (a) the potential at point $A ;$ (b) the potential at point $B ;$ (c) the work done by the electric field on a charge of $2.50 \mathrm{nC}$ that travels from
point $B$ to point $A$.

Key Concepts

Electric Potential of a Point Charge

This concept involves the idea that each point charge creates an electric potential in the space around it, defined by the equation V = kq/r, where k is Coulomb's constant, q is the charge, and r is the distance from the charge. The potential is defined with reference to a point where it is considered to be zero, typically at infinity.

Superposition Principle in Electrostatics

The superposition principle states that the total electric potential at any point due to several point charges is the algebraic summation of the potentials due to each individual charge. This allows for the determination of the net potential in a system with multiple charges by adding the contributions from each charge.

Work Done by an Electric Field

In an electric field, the work done on a charge when it moves from one point to another is related to the change in electric potential, given by W = q(V_initial - V_final). This concept connects the energy transfer in the field to the displacement and the electric potentials at the starting and ending points.

Transcript

00:01 In the given problem there are the two charges plus q1 and minus q2 which are having the values q1 is plus 2 .40 nanoculum and q2 is minus 6 .50 nanoculum and q2 is minus 6 .50 nanoculum the gap between them is given as 0 .10 meter r is equal to then then there is a point a exactly in the middle of the two charges so these distances will be 0 .05 meter each and another point b whose distance from this q1 is 0 .080 meter and from q2 this is 0 .060 meter.

01:36 In the first part of the problem we have to find the potential at point a.

01:48 And that will be because of the two charges it will be given by k into q1 divided by distance 0 by 0 5 meter plus k u2 divided by the same distance 0 .05 meter taking this k whose value is 9 into 10 dash to the part 9 and taking it common leaving behind the value of k1 which is 2 .40 nano and for nano this is minus 9 10 dash per minus 9 divided by r which is 0 .05 we have already seen then plus it will be minus 6 .50 nano again so again 10 dash per minus 9 having the same distance 0 .5 0 .05 meter now taking this and dash bar minus 9 also as a common out then 2 .40 divided by 0 .05 will come out to be 30 minus 6 .50 divided by 0 .05 it will come out to be 10028 .3 sorry we have done the wrong calculations here actually it will here come out to be 48 and here the second term will come out to be 130 and that will be in volt canceling this 10 tish per plus 9 with the 10 dash for minus 9 finally we get the potential at point a to be equal to minus 738 volt which is the answer for the first part of the problem.

04:00 Now, in the second part of the problem, we have to find now the potential at the point b.

04:09 And again, that is cause of the two charges, one q1 and another q2.

04:15 So that is k into q1 by the distance 0 .08 meter plus k.

04:26 Q2 by 0 .06 meter.

04:31 So, taking this k as a common out, which is 9 into 10 dash par minus 9, sorry, plus 9, and then putting the value of q1, which is 2 .40 into 10 dashpar minus 9 divided by 0 .08 plus the value of q2, which is minus 6 .50...

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