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I wanted to know that can we find out the potential difference across ends of a metallic rod having cross-sectional area A, length of rod l cm and mass m kg if the Force of F N is acting on it along its length ?

my idea: we will find force as F = -(charge)*(dV/dr). But what next?

justaboi
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  • Why is the force F acting on the rod? What is the nature of this force? – nasu Mar 05 '22 at 06:00
  • @nasu force is acting along the length and nature of force is assumed constant. – justaboi Mar 06 '22 at 05:35
  • By "nature of force" I mean why there is force at all? What kind of force is this? A piece a wire does not experience a force unless there is some interaction with something else. – nasu Mar 06 '22 at 19:15
  • @nasu i believe the force given in question is assumed to be conservative force. – justaboi Mar 12 '22 at 07:16
  • @justaboy But what is producing that force? – nasu Mar 12 '22 at 12:32

1 Answers1

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In order to determine the potential difference between the two ends of the rod you need to know the resistance $R$ of the rod and the current $I$ in the rod. To determine $R$ you need the electrical resistivity $\rho$ of the metal, length and cross sectional area of the rod. Then apply Ohm's law $V=IR$. The mass of the rod is irrelevant. So

$$R=\rho\frac{L}{A}$$ $$V=IR=\rho \frac{IL}{A}$$

What you are attempting to do ends up as circular reasoning. The potential difference $V$ between the ends of the rod is the work $W$ required per unit charge $Q$ to move the charge a distance $d$, or

$$V=\frac{W}{Q}=\frac{Fd}{Q}\tag{1}$$

The magnitude of the force on the charge equals the product of the charge and the magnitude of the electric field $E$ in the rod, or

$$F=QE\tag{2}$$

Substituting (2) into (1)

$$V=\frac{QEd}{Q}=Ed\tag{3}$$

The electric field is the voltage gradient in the rod, assumed uniform, or

$$E=\frac{V}{d}\tag{4}$$

Substituting (4) into (3)

$$V=\frac{V}{d}d=V\tag{5}$$

Which provides no information on the potential difference between the ends of the rod based on the rods physical and electrical characteristics.

Hope this helps.

Bob D
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