Question

Question 2.14: Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field: (a) at the mid-point of the line joining the two charges, and (b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-88

Answer 1

Let two charges $1.5\mu C$ and $2.5\mu C$ are placed on line of points A and B . at the midpoint C f line joining two charges , electric potential is $V_C=V_{CA}+V_{CB}\\\\or,V_C=\frac{1}{4\pi\epsilon_0}\left[\begin{array}{c}\frac{q_A}{r_{CA}}+\frac{q_B}{r_{CB}}\end{array}\right]\\\\=9\times10^9\left[\begin{array}{c}\frac{1.5}{15}+\frac{2.5}{15}\end{array}\right]\times\frac{10^{-6}}{10^{-2}}\\\\or,V_C=2.4\times10^5V$

and electric field at point C is $E_C=E_{CB}-E_{CA}$ [As $E_{CA}>E_{CB}$ and they are directed opposite to each other]

$E_C=\frac{1}{4\pi\epsilon_0}\left[\begin{array}{c}\frac{q_B}{r_{CB}^2}-\frac{q_A}{r_{CA}^2}\end{array}\right]\\\\=9\times10^9\left[\begin{array}{c}\frac{2.5}{15^2}-\frac{1.5}{15^2}\end{array}\right]\times\frac{10^{-6}C}{10^{-4}m^2}\\\\or, E_C=4\times10^5V/m$
and direction ofn $E_C$ from C to A.

(b) see attachment,

Answer 2

Answer:

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