Respuesta :
Answer:
Explanation:
Given
for [tex]\theta=10^{\circ}[/tex]
Sphere are [tex]d=40\ cm[/tex]
when sphere [tex]d_2=10\ cm[/tex] apart suppose deflection is [tex]\theta _2[/tex]
We know
[tex]F=k_t\cdot \theta [/tex]
Where F=force between charged particle
[tex]\theta =[/tex]Deflection
[tex]F=\frac{kQ_1Q_2}{r^2}=k_t\cdot \theta [/tex]
[tex]\theta =\frac{k}{k_t}\times \frac{Q_1Q_2}{r^2}----1[/tex]
thus [tex]\theta \propto \frac{1}{r^2}[/tex]
for [tex]\theta _2[/tex]
[tex]\frac{\theta _1}{\theta _2}=(\frac{r_2}{r_1})^2[/tex]
[tex]\theta _2=16\times \theta _1[/tex]
[tex]\theta _2=160^{\circ}[/tex]
(b)for [tex]10^{\circ}[/tex] deflection Potential [tex]v_1=8\ kV[/tex]
Electric Potential is [tex]V=\frac{kQ}{r}[/tex]
[tex]Q=\frac{V\cdot r}{k}[/tex]
where V=voltage
k=constant
r=distance between charges
Put value of Q in equation 1
[tex]\theta =\frac{k}{k_t}\times \frac{V^2r^2}{k^2}[/tex]
[tex]\theta =\frac{V^2r^2}{k\cdot k_t}[/tex]
thus [tex]\theta \propto V^2[/tex]
therefore
[tex]\frac{\theta _1}{\theta _2}=(\frac{V_1}{V_2})^2[/tex]
[tex]\frac{10}{\theta _2}=(\frac{8}{4})^2[/tex]
[tex]\theta _2=\frac{10}{4}=2.5^{\circ}[/tex]
