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

Step-by-step 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=(kQ_1Q_2)/(r^2)=k_t\cdot \theta`

`\theta =(k)/(k_t)* (Q_1Q_2)/(r^2)----1`

thus
`\theta \propto (1)/(r^2)`

for
`\theta _2`

`(\theta _1)/(\theta _2)=((r_2)/(r_1))^2`

`\theta _2=16* \theta _1`

`\theta _2=160^(\circ)`

(b)for
`10^(\circ)` deflection Potential
`v_1=8\ kV`

Electric Potential is
`V=(kQ)/(r)`

`Q=(V\cdot r)/(k)`

where V=voltage

k=constant

r=distance between charges

Put value of Q in equation 1

`\theta =(k)/(k_t)* (V^2r^2)/(k^2)`

`\theta =(V^2r^2)/(k\cdot k_t)`

thus
`\theta \propto V^2`

therefore

`(\theta _1)/(\theta _2)=((V_1)/(V_2))^2`

`(10)/(\theta _2)=((8)/(4))^2`

`\theta _2=(10)/(4)=2.5^(\circ)`