Respuesta :
Answer:
[tex]200\times 10^{-6}j[/tex]
Explanation:
We have given the radius of first sphere is 10 cm and radius of second sphere is 20 cm
So the potential of first sphere will be greater than the potential of the second sphere, so charge will flow from first sphere to second sphere
Let q charge is flow from first sphere to second sphere and then potential become same
So [tex]V=\frac{K(100-q)}{r_1}=\frac{K\times 100}{r_2}[/tex]
200-100=2q+q
[tex]q=\frac{100}{3}=33.33nC[/tex]
So [tex]V=\frac{K(100-q)}{r_1}=\frac{9\times 10^{9}\times (100-33.33)\times 10^{-9}}{10\times 10^{-2}}=6003V[/tex]
We know that potential energy U=qV[tex]=33.33\times 10^{-9}\times 6003=200\times 10^{-6}j[/tex]
Answer:
The electric potential energy between the two charged spheres is [tex]199.9\times10^{-6}\ J[/tex]
Explanation:
Given that,
Radius of first sphere [tex]R_{1}=10\ cm[/tex]
Radius of second sphere [tex]R_{2}=10\ cm[/tex]
Charge Q= 100 nC
We know charge flows through higher potential to lower potential.
Using formula of potential
[tex]V=\dfrac{k(Q-q)}{R_{1}}[/tex]...(I)
[tex]V=\dfrac{k(Q+q)}{R_{2}}[/tex]...(II)
From equation (I) and (II)
[tex]\dfrac{k(Q-q)}{R_{1}}=\dfrac{k(Q+q)}{R_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{(100-q)}{10\times10^{2}}=\dfrac{(100+q)}{20\times10^{-2}}[/tex]
[tex](100-q)\times20\times10^{-2}=(100+q)\times10\times10^{-2}[/tex]
[tex]q=\dfrac{1000}{30}[/tex]
[tex]q=\dfrac{100}{3}[/tex]
[tex]q=33.33\ nC[/tex]
So, the potential at R₁ and R₂
Using formula of potential
[tex]V=\dfrac{k(Q-q)}{R_{1}}[/tex]
Put the value into the formula
[tex]V=\dfrac{9\times10^{9}(100-33.33)\times10^{-9}}{10\times10^{-2}}[/tex]
[tex]V=6000.3\ Volt[/tex]
We need to calculate the electric potential energy between the two charged spheres
Using formula of the electric potential energy
[tex]U=qV[/tex]
[tex]U=33.33\times10^{-9}\times6.0003\times10^{3}[/tex]
[tex]U=199.9\times10^{-6}\ J[/tex]
Hence, The electric potential energy between the two charged spheres is [tex]199.9\times10^{-6}\ J[/tex]
