0
$\begingroup$

In an ideal battery, the battery forces are equal in magnitude to the electric forces (the ones that are present when battery is not connected externally) and thus we conclude that E.M.F. ($\varepsilon$) of an ideal battery equals the potential difference between its terminals when terminals are not connected externally

$$\varepsilon=\frac{F_bd}{q}$$ $$F_b=qE_*$$ $$\varepsilon=E_*d=V$$

$F_b=$ battery force, $d=$ distance between terminals, $V=$ petential difference between the terminals, $E_*=$ electric field between terminals when not connected externally

Now when we connect the terminals externally the electric forces tends to decrease(due to flow of charges externally) and $F_b$ dominates and due to this difference the charges flow inside the battery too. Now my question is if the magnitude of electric forces becomes less than the $F_b$ (which is same as earlier) then $V(=Ed)$ should also decrease (since $E$ decreases). Then why do we say and eventually do calculations by considering that the potential difference between the terminals of an ideal battery remains equal to its $\varepsilon$ even if it is connected externally.

Since I'm a new learner so I may be wrong somewhere in the understanding of battery. Please correct me if I'm wrong so that I can better understand the concepts

Improve this question
$\endgroup$
3
  • $\begingroup$ Why does E decrease? $\endgroup$
    – Stack Exchange Supports Israel
    Commented Jul 14, 2020 at 15:24
  • $\begingroup$ @user253751 since charges flow externally thus reducing $E$ inside battery and this make $F_b$ relatively stronger and there’s current inside battery too. $\endgroup$
    – abcxyz
    Commented Jul 14, 2020 at 16:25
  • $\begingroup$ Do you know what "ideal" means? $\endgroup$
    – Stack Exchange Supports Israel
    Commented Jul 14, 2020 at 17:16

0

Reset to default