Question

A)If your torsion balance deflects to an angle of 10° when your two spheres are 40cm apart, what angle will it deflect to when the spheres are 10cm apart?
b)If your torsion balance deflects to an angle of 10° when your two spheres are 20cm apart, and charged to a potential of 8 kV, what angle will it deflect to when the spheres are charged to a voltage of 4kV?

Answer 1

Answer:

Explanation:

Given

for $\theta=10^{\circ}$

Sphere are $d=40\ cm$

when sphere $d_2=10\ cm$ apart suppose deflection is $\theta _2$

We know

$F=k_t\cdot \theta$

Where F=force between charged particle

$\theta =$Deflection

$F=\frac{kQ_1Q_2}{r^2}=k_t\cdot \theta$

$\theta =\frac{k}{k_t}\times \frac{Q_1Q_2}{r^2}----1$

thus $\theta \propto \frac{1}{r^2}$

for $\theta _2$

$\frac{\theta _1}{\theta _2}=(\frac{r_2}{r_1})^2$

$\theta _2=16\times \theta _1$

$\theta _2=160^{\circ}$

(b)for $10^{\circ}$ deflection Potential $v_1=8\ kV$

Electric Potential is $V=\frac{kQ}{r}$

$Q=\frac{V\cdot r}{k}$

where V=voltage

k=constant

r=distance between charges

Put value of Q in equation 1

$\theta =\frac{k}{k_t}\times \frac{V^2r^2}{k^2}$

$\theta =\frac{V^2r^2}{k\cdot k_t}$

thus $\theta \propto V^2$

therefore

$\frac{\theta _1}{\theta _2}=(\frac{V_1}{V_2})^2$

$\frac{10}{\theta _2}=(\frac{8}{4})^2$

$\theta _2=\frac{10}{4}=2.5^{\circ}$