Two point charges +q and -q lie along the y axis separated by a distance of 2d. Charge -q lies at...

Question:

Two point charges +q and -q lie along the y axis separated by a distance of 2d. Charge -q lies at y = +d and charge +q lies at y = -d.

a. Determine the net electric field at point A (x = a, y = 0) and point B ( x = 0, y = -b).

b. Determine the electric potential at point A (x = a, y = 0) and point B ( x = 0, y = -b).

Electric Field and Potential of Point Charge Distribution

The electric field due to a point charge is defined by,

{eq}\displaystyle \vec E=\frac{kq}{r^2}\hat r {/eq}

The electric potential is,

{eq}\displaystyle V=\frac{kq}{r} {/eq}

Where,

  • {eq}k=8.99\times10^9\ Nm^2/C {/eq} is the Coulombic constant,
  • {eq}q {/eq} is the magnitude of the charge,
  • {eq}r {/eq} is the distance from the point charge and
  • {eq}\hat r {/eq} is the unit vector from the point charge to the field point.

Electric field is a vector and potential is a scalar.

Answer and Explanation:

Let {eq}q_1=q\\ q_2=-q {/eq}

a)

For point A the electric field due to first charge is,

{eq}\displaystyle \vec{E_1}=\frac{kq_1}{(a^2+d^2)}\frac{a\hat x+d\hat y}{\sqrt{a^2+d^2}} {/eq}

Field due to second charge is,

{eq}\displaystyle \vec{E_2}=\frac{kq_2}{(a^2+d^2)}\frac{a\hat x-d\hat y}{\sqrt{a^2+d^2}} {/eq}

Then, resultant field is,

{eq}\displaystyle \vec {E_A}=\vec{E_1}+\vec{E_2}=\frac{kq}{(a^2+d^2)^{3/2}}(a\hat x+d\hat y-a\hat x+d\hat y)=\frac{2kqa\hat x}{(a^2+d^2)^{3/2}} {/eq}

For point B the field due to first and second charges are,

{eq}\displaystyle \vec{E_1}=\frac{kq_1}{(d-b)^2}\hat y\\ \displaystyle \vec{E_2}=\frac{-kq_2}{(d+b)^2}\hat y {/eq}

Therefore, resultant field at B is,

{eq}\displaystyle {E_B}=\vec{E_1}+\vec{E_2}=kq(\frac{\hat y}{(d-b)^2}+\frac{\hat y}{(d+b)^2})=kq(\frac{1}{(d-b)^2}+\frac{1}{(d+b)^2}) {/eq}

b)

Potential at A is,

{eq}\displaystyle V_A=\frac{kq_1}{\sqrt{a^2+d^2}}+\frac{kq_2}{\sqrt{a^2+d^2}}=0 {/eq}

Potential at B is,

{eq}\displaystyle V_B=\frac{kq_1}{d-b}+\frac{kq_2}{d+b}=kq(\frac{1}{d-b}-\frac{1}{d+b}) {/eq}


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