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If two Hollow Spheres with radius $r_1$ and $r_2$ have charge $q_1$ and $-q_2$ in them respectively(charges are evenly distributed in the surfaces).The spheres are separated by a distance d .Now We connect a conducting wire from one sphere to another.Then What will be the potential difference between the two ends of that wire and what will be the amount of current flow if the resistance is R in that wire(lets assume the amount of charges in each sphere always remain constant i.e. $q_1$ and $-q_2$)? Can we evaluate these things mathematically i.e. without using the voltmeter or ammeter?

Like, i was thinking of using $\frac{kq}{r} $ formula for calculating the potential at each end and then $V_A-V_B=IR$ to Calculate the amount of Current flow.Is that a right approach?

MD Hossain
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1 Answers1

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The potential difference

Finding the potential difference is a bit more complicated than you suggested, if I interpreted the meaning of $k$, $q$ and $r$ in your formula correctly.

Before connecting the spheres, they effectively form a capacitor with some capacitance $C$. If $r_1 = r_2$, you can use a formula from wikipedia to calculate $C$, which then gives you the voltage (potential difference) as $$ V = \frac 1C \frac{q_1+q_2}{2}~. $$ If $r_1 \neq r_2$, you will have to solve the Poisson equation, $$ \Delta \phi(\vec r) = - \frac{\rho(\vec r)}{\varepsilon_0}~, $$ where $\phi$ is the electrical potential, $\rho$ is the charge density, $\vec r$ is the position, $\Delta$ is the Laplacian and $\varepsilon_0$ is the dielectric constant of the vacuum. The total charge on each of the spheres serves as boundary condition for the solution. As soon as a solution $\phi(\vec r)$ is found, the potential difference can directly be calculated.

The electric current

Once the voltage is known, the current is, by Ohm's law, simply $$ I = \frac VR~, $$ just as you stated. However this only holds immediately upon connection of the spheres. Afterwards, it makes little sense to assume the charges remain constant, because electric current is just a flow of charge which will eventually cause equal charge density on both spheres and the wire. This is nothing else than the discharge process of a capacitor, which can be modeled quite easily:

  • Let $t$ be the time and $t = 0$ the point at which the spheres are connected. Now the current $I(t)$, the charge $Q(t)$ and the voltage $V(t)$ vary over time.
  • Using Ohm's law, we have $V(t) = I(t) R$ and per definition of the capacitance it holds $V(t) = Q(t) / C$. This means $$ \frac{Q(t)}{C} = I(t) R \qquad \Leftrightarrow \qquad \frac{Q(t)}{C} - I(t) R = 0~. $$
  • The current $I(t)$ is a change (the derivative) of charge, so $I(t) = \partial_t Q(t)$ and we have $$ \frac{Q(t)}{C} - R \partial_t Q(t) = 0 \qquad \Leftrightarrow \qquad Q(t) - RC \partial_t Q(t) = 0~. $$
  • The solution of this first-order linear ordinary differential equation is $$ Q(t) = Q(0) e^{-RCt}~, $$ which is a parametrization of exponential decay.
sim0
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