Answer to Question #124041 in Electricity and Magnetism for alexander

Question

4.

A.

The diagram at right shows two charged objects that are separated by a distance L. Object #1 bears a

charge +7Q and object #2 a charge of ‒4Q. A point in space, P, is located a distance y directly above object

#2.

Write an expression for the y-component of the net electric field at point P. (xmm)

B. Write an expression for the net electric potential at point P.

C. Rederive your expression for the y-component of the electric field from part A, this time by taking the negative gradient of the electric potential expression you derived in part B. Note: little credit will be awarded for an answer that is not fully explained.

Answer 1

As per the given question,

magnitude of charge 1 on one end $(Q_1)=7Q$

magnitude of charge 2 on other end $(Q_2)=-4Q$

these two charges 1 and 2 are situated at a distance of L to each other.

point p is placed at a distance y, from the middle of line joining charges.

Net electric field at the point P,

$\Rightarrow E_y=\frac{7Q}{4\pi \epsilon_o ((\frac{L}{2})^2+y^2)}\cos\theta_1+\frac{4Q}{4\pi \epsilon_o ((\frac{L}{2})^2+y^2)}\cos\theta_2$

$\Rightarrow E_y =\frac{56Q}{4\pi \epsilon_o(L^2+4y^2)}\frac{y}{\sqrt{ ((L^2)^2+y^2)}}+\frac{8Q}{4\pi \epsilon_o ((L^2)^2+y^2)}\frac{y}{\sqrt{ ((L^2)^2+y^2)}}$

$\Rightarrow E_y= \frac{64Q}{4\pi \epsilon_o ((L^2)^2+y^2)}\frac{y}{\sqrt{ ((L^2)^2+y^2)}} \hat{j}N/coulamb$

Net electric field along the x axis,

$\Rightarrow E_x=\frac{7Q}{4\pi \epsilon_o ((\frac{L}{2})^2+y^2)}\sin\theta_1+\frac{4Q}{4\pi \epsilon_o ((\frac{L}{2})^2+y^2)}\sin\theta_2$

$\Rightarrow E_x =\frac{28Q}{4\pi \epsilon_o(L^2+4y^2)}\frac{L}{\sqrt{ ((L^2)^2+y^2)}}+\frac{4Q}{4\pi \epsilon_o ((L^2)^2+y^2)}\frac{L}{\sqrt{ ((L^2)^2+y^2)}}$

$\Rightarrow E_x=\frac{32Q}{4\pi \epsilon_o ((L^2)^2+y^2)}\frac{L}{\sqrt{ ((L^2)^2+y^2)}} \hat{i}$

Hence, net electric field

$E=E_x\hat{i}+E_y\hat{j}$

$\Rightarrow E_{net}=\frac{32Q}{4\pi \epsilon_o ((L^2)^2+y^2)}\frac{L}{\sqrt{ ((L^2)^2+y^2)}} \hat{i}+ \frac{64Q}{4\pi \epsilon_o ((L^2)^2+y^2)}\frac{y}{\sqrt{ ((L^2)^2+y^2)}} \hat{j}$

Net potential at the point P,

$\Rightarrow V=V_1+V_2$

$\Rightarrow V=\frac{7Q}{4\pi \epsilon \sqrt{ ((L^2)^2+y^2)}}-\frac{4Q}{4\pi \epsilon \sqrt{ ((L^2)^2+y^2)}}$

$\Rightarrow V=\frac{3Q}{4\pi \epsilon \sqrt{ ((L^2)^2+y^2)}}$

As the mentioned diagram is not given in the question, hence the situation telling in the question is not solvable.