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Why the work done by a battery is Q*V where V is emf of battery and Q is charge that is made to flow in circuit?please explain detail? explain and write the formulas

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    $\begingroup$ One definition of the SI unit volt is is the potential difference between two points that will impart one joule of energy per coulomb of charge that passes through it. $\endgroup$ – gandalf61 May 21 at 9:23
  • $\begingroup$ Well according to E=QV, voltage is just the energy per coulomb. Multiply by the number of coulombs that carry that energy you get just the energy $\endgroup$ – Ubaid Hassan May 21 at 13:22
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As a result of chemical reactions in the battery, if no circuit is connected to the battery charges will accumulate at the electrodes. The battery will have an open circuit voltage (its emf) between its terminals, and an electric field, $E$, inside the battery between anode and cathode.

When a circuit is connected, using conventional current as the flow of positive charge, $Q$, the charge will go from the positive terminal to the negative terminal by way of the circuit, losing potential energy. In order to move the returning charge from its negative terminal to its positive terminal, the field exerts a force of $QE$. In so doing the battery does work moving positive charge from its negative terminal to its positive terminal (increasing the potential energy of the positive charge) equal to $QEd$ where $d$ is the distance in the battery between the electrodes.

Voltage, or electrical potential, is defined as the work per unit charge required to move the charge between two points, or $V=\frac{QEd}{Q}=Ed$.

Therefore the work done by the battery is $QV$.

Hope this helps

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  • $\begingroup$ "A battery produces a constant electric field, call it E, in the circuit to which it is connected." Are you stating this as a fact for all circuits, or setting it as a condition for a specific example circuit? A constant E would mean that the voltage is changing linearly throughout the size of the circuit, and that's not the case in most circuits. $\endgroup$ – Bill N May 21 at 13:33
  • $\begingroup$ @BillN You are right. The electric field will be determined around the conductor depending on the change in potential along the conductor. What I should have said it produces an electric field inside the battery due to accumulation of charges at the electrodes. I will edit to correct. Thanks. $\endgroup$ – Bob D May 21 at 13:44
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The work done by a force ${\bf F}$ moving a particle through the displacement $d {\bf r}$ is $\bf{F \cdot} d {\bf r}$. For a charge $Q$ in an electric field ${\bf E}$ the force acting on the charge is ${\bf F} = Q {\bf E}$. The electric field defined in terms of the scalar potential $V$ is ${\bf E} = - {\bf \nabla} \,V$ so that the work done is $ - Q\bf{\nabla} V \bf{\cdot} d {\bf r}$. To determine the total work done integrate around the circuit, which gives (apart from a sign) $Q V$.

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