SATHEE: Electrostatics Question 689

Electrostatics Question 689

Question: Two pith balls carrying equal charges are suspended from a common point by strings of equal the balls now become

Options:

A) $(\frac{r}{\sqrt[3]{2}})$

B) $(\frac{2 r}{\sqrt{3}})$

C) $(\frac{2 r}{3})$

D) ${(\frac{r}{\sqrt{2}})}^{2}$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $\tan θ = \frac{F_{e}}{m g} \Rightarrow \frac{r / 2}{y} = \frac{\frac{k q^{2}}{r^{2}}}{m g}$

[ $∵ F = \frac{k q^{2}}{r^{2}}$

from coulomb’s law] $\Rightarrow r^{3} \propto y \Rightarrow r {^{'}}^{3} \propto \frac{y}{2} \Rightarrow \frac{r^{'}}{r} = \frac{1}{2^{1 / 3}} \Rightarrow r^{'} = \frac{r}{\sqrt[3]{2}}$

Electrostatics Question 689

Question: Two pith balls carrying equal charges are suspended from a common point by strings of equal the balls now become

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Electrostatics Question 689

Question: Two pith balls carrying equal charges are suspended from a common point by strings of equal the balls now become

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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