Two point charges -5 mu C and +5 mu C are pla | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the charges be $q_A = -5 \mu C$ and $q_B = +5 \mu C$. Let the distance $AB = d = 5 \ cm$. Let point $C$ be at a distance $x$ from $A$ and $y$

Vedclass · Accepted Answer

$AC = BC$

Vedclass · Answer

(A) Let the charges be $q_A = -5 \mu C$ and $q_B = +5 \mu C$. Let the distance $AB = d = 5 \ cm$. Let point $C$ be at a distance $x$ from $A$ and $y$ from $B$. For the resultant electric field at $C$ to be parallel to the line $AB$,the vertical components of the electric fields produced by $q_A$ and $q_B$ must cancel each other out.Let $	heta_A$ and $	heta_B$ be the angles made by the vectors $\vec{E}_A$ and $\vec{E}_B$ with the line $AB$. For the vertical components to cancel,we must have $E_A \sin 	heta_A = E_B \sin 	heta_B$.Since the magnitudes of the charges are equal $(|q_A| = |q_B| = q)$,the field magnitudes are $E_A = \frac{kq}{x^2}$ and $E_B = \frac{kq}{y^2}$.Thus,$\frac{kq}{x^2} \sin 	heta_A = \frac{kq}{y^2} \sin 	heta_B$. From the geometry of the triangle $ABC$,$\sin 	heta_A = \frac{h}{x}$ and $\sin 	heta_B = \frac{h}{y}$,where $h$ is the perpendicular distance of $C$ from line $AB$.Substituting these,we get $\frac{kq}{x^2} \cdot \frac{h}{x} = \frac{kq}{y^2} \cdot \frac{h}{y}$,which simplifies to $\frac{1}{x^3} = \frac{1}{y^3}$,implying $x = y$.Therefore,$AC = BC$.

Two point charges $-5 \mu C$ and $+5 \mu C$ are placed at $A$ and $B$ respectively on a straight line separated by $5 \ cm$. At a point $C$,if the resultant electric field is parallel to the line joining the points $A$ and $B$,then:

Similar Questions

Four point charges are placed at the vertices of a regular hexagon as shown in the figure below. What will be the net electric field at the centre?

The electric field at the center of a semi-circle of radius $a$ and linear charge density $\lambda$ is given by:

The maximum value of the electric field on the axis of a charged ring having charge $Q$ and radius $R$ is:

Find out the electric field intensity at point $A(1, 0, 2)$ due to a point charge $-20\,\mu C$ situated at point $B(0, 2, 1)$.

$A$ spherical conductor of radius $12 \;cm$ has a charge of $1.6 \times 10^{-7} \;C$ distributed uniformly on its surface. What is the electric field
$(a)$ inside the sphere
$(b)$ just outside the sphere
$(c)$ at a point $18\; cm$ from the centre of the sphere?

Vedclass Test Series

Exam Paper Generator

Online Exam Module