Two charges of 10 microcolumb and -10 microcolumb are placed at a points A and B separated by distance of10 cm. find the electric field at a point at a p on the perpendicular bisector of AB at a distance of 12cm from its middle point ? a) 16.4×10⁶N/c b)28.4×10⁶N/c c)8.2×10⁶N/c d) 4.1×10⁶N/c?

Question

Answer

Answer

Answer

Electric Field Due to a Point Charge:
The electric field due to a point charge Q at a distance r from the charge is given by Coulomb's law:

E = k * Q / r^2

where E is the electric field, k is the electrostatic constant (9 * 10^9 N m^2 / C^2), Q is the charge, and r is the distance.

Electric Field at the Midpoint of AB:
To find the electric field at the midpoint of AB, we can use the principle of superposition. Since the charges at points A and B are equal in magnitude and opposite in sign, they will produce electric fields of equal magnitude and opposite in direction at the midpoint.

The electric field due to each charge at the midpoint will be:

E1 = k * Q / (0.05^2) = 20000 * k * Q N/C (pointing towards A)
E2 = k * (-Q) / (0.05^2) = -20000 * k * Q N/C (pointing towards B)

The net electric field at the midpoint is the vector sum of these two electric fields:

E_net = E1 + E2 = 0

Since the net electric field at the midpoint is zero, the charges at points A and B cancel each other out.

Electric Field at Point P on the Perpendicular Bisector:
To find the electric field at point P on the perpendicular bisector of AB, we can consider the electric fields due to each charge separately and then add them vectorially.

The electric field due to the charge at point A at point P is given by:

E1 = k * Q / (0.16^2) = 3125 * k * Q N/C (pointing towards A)

The electric field due to the charge at point B at point P is given by:

E2 = k * (-Q) / (0.04^2) = -62500 * k * Q N/C (pointing towards B)

The net electric field at point P is the vector sum of these two electric fields:

E_net = E1 + E2 = (3125 - 62500) * k * Q N/C = -59375 * k * Q N/C

The magnitude of the net electric field at point P is:

|E_net| = 59375 * k * Q N/C

Substituting the values of k (9 * 10^9 N m^2 / C^2) and Q (10 * 10^-6 C) into the equation, we get:

|E_net| = 59375 * (9 * 10^9) * (10 * 10^-6) N/C = 16.4 * 10^6 N/C

Therefore, the electric field at point P on the perpendicular bisector of AB is 16.4 * 10^6 N/C.

Answer: a) 16.4×10⁶N/c

Similar NEET Doubts

Two point charges of 1μC and -1μC are separated by a distance of 100 Å. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be

Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance will experience maximum coulomb force when

Two wires AB & CD, each Im length, carry a total charge of0.2 microcoulomb each and are placed The ends B & C are separated 1m distance. Determine the value of electric intensity at the mid point P in terms of unit vector i and j.?

Two equal charge A and B are 'R' distance apart. A third point charge placed on the perpendicular to the bisector at a distance D from the center will experience maximum electrostatic force when?? (1) D= R/2√2 (2) D=R/√2?

A point B lies on the perpendicular bisector of an electric dipole of dipole moment p the distance of B from the dipole is r. The electric intensity at B is proportional to 1)r 2)1/r square 3)1/r cube 4)r square Answer is option 4 can you explain why?

Top Courses for NEETView all

Biology Class 11

Daily Test for NEET Preparation

NEET Mock Test Series 2025

Topic-wise MCQ Tests for NEET

Physics Class 11

Top Courses for NEET

Top Courses for NEET

Suggested Free Tests

Related Content

About NEET

NEET Preparation Material

How to Prepare for NEET Exam?

NCERT Based Study Material

NEET Mock Tests

Other Popular Exams

Company