Two small spheres,each having an equal positi | vedclass

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

Question

(A) In the absence of gravity,the only force acting on each sphere is the electrostatic repulsive force exerted by the other sphere.Since the spheres

Vedclass · Accepted Answer

$\frac{Q^2}{16 \pi \varepsilon_0 L^2}$

Vedclass · Answer

(A) In the absence of gravity,the only force acting on each sphere is the electrostatic repulsive force exerted by the other sphere.Since the spheres are held in a horizontal position,the distance between the two spheres is $2L$.According to Coulomb's law,the electrostatic force $F$ between the two charges is given by:$F = \frac{1}{4 \pi \varepsilon_0} \frac{Q \cdot Q}{(2L)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q^2}{4L^2} = \frac{Q^2}{16 \pi \varepsilon_0 L^2}$.Since the spheres are in equilibrium,the tension $T$ in each string must balance this electrostatic force.Therefore,$T = F = \frac{Q^2}{16 \pi \varepsilon_0 L^2}$.

Similar Questions

If $a > 0$ and $b^2 - 4ac = 0$,then the curve $y = ax^2 + bx + c$

If $\log (1+x) - \frac{2x}{2+x}$ is increasing,then

Assign $A, B, C, D$ from the given type of reaction.
$NaCl + AgNO_3 \longrightarrow AgCl \downarrow + NaNO_3$

$A$ beam of light of wavelength $600\, nm$ from a distant source falls on a single slit $1\, mm$ wide and the resulting diffraction pattern is observed on a screen $2\, m$ away. The distance between the first dark fringes on either side of the central bright fringe is

Vedclass Test Series

Exam Paper Generator

Online Exam Module