When the charging is finished what will be the charge on an ideal parallel plate capacitor ? To be specific what are the charges on different faces of the 2 plates ? enter image description here

All the books say, after 100% charging, the final charge on capacitor is $$ Q = CV $$

But as far as I think the net charge on the capacitor is 0. So what do we actually mean when we say that the capacitor is having a charge $Q$. I mean, what would the charges on the plates be and most importantly why ?

A Brief explanation on how these charges are coming up on these plates will be highly appreciated. Also, I have searched for it in not only books but also Wikipedia and some SE related sites like Hyperphysics etc. But whatever information I have gained is confusing me as I find some answers contradicting. This is my last attempt to understand the concept.

I have checked these, but were not really helpful : Charging a capacitor (terminals) , Charging a capacitor

Thank you.

  • $\begingroup$ Consider: suppose one side of the cap is tied to ground, where "ground" is considered an infinite supply/drain of charge. Now what charge is "inside" the cap? $\endgroup$ – Carl Witthoft Nov 5 '15 at 13:29

By capacitor charge is meant the absolute value of the charge on each capacitor plate: $ \mid Q \mid$. If the battery generates the potential difference $V$ and you connect the capacitor to the battery through a conducting wire, as shown in your picture, once the equilibrium is reached each plate of the capacitor will have a charge $ Q = CV $, where $C$ is the capacitor capacitance.

In your case: $ q_2 = \mid Q \mid $, since that plate is connected to the battery positive pole and $ q_3 = - \mid Q \mid $, since that plate is connected to the negative pole. Finally $ q_1 = q_4 = 0 $ because the electric field generated by the battery will drive the electrons to the positive pole and as far as possible from the negative one.

So as you said the net charge on the system will be zero. The charging process can happen because the battery keeps a constant potential and thus an electric field in the conducting materials (wire and capacitor plate).

  • $\begingroup$ If i am going to remove the battery, what is going to happen to the distribution of charges ? $\endgroup$ – Shubham Nov 5 '15 at 15:35
  • $\begingroup$ For and ideal capacitor the charge on the plates will remain constant. Note that $ Q = CV $ also means that the plates charged with $ + \mid Q \mid $ and $ - \mid Q \mid $ will generate a potential equal to $ V $. So even in the battery is removed, the capacitor will still generate the potential $ V $. That's why capacitors can be used to store electrostatic energy. $\endgroup$ – NNec Nov 5 '15 at 15:54

enter image description here

$$V = {v_r}(t) + {v_c}(t) = i(t)R + {1 \over C}\int\limits_{{t_0}}^t {i(\tau )d} \tau $$ Differentiating wrt t: $$RC{{di(t)} \over {dt}} + i(t) = 0$$ Solving differential equation: $$I(t) = {V \over R}{e^{{t \over {{\tau _0}}}}}$$ From this equation we notice initially at t = 0 $$I = {V \over R}$$

As time is increasing current starts decreasing until at $t = \infty $ the current in the circuit becomes 0 and at this time $t = \infty $ the capacitor gets 100% charged. But in reality t never reaches $\infty $ and thus, at the limit point(from calculus) the charge never goes |Q| exactly but it certainly remains closest to |Q|, $\forall $ (t$ \le \infty $). By |Q| I mean charge on the single plate of the capacitor.

Drawing parallelism with the diagram in question, suppose the dielectric slab completely fills the spaces between plates a charge ${q_1}$=+Q is given to the positive plate and ${q_4}$=-Q to the negative plate of the capacitor. The electric field polarizes the dielectric so that the induced charges ${q_2} = - {Q_p}$ and ${q_3} = + {Q_p}$ appear on the faces of the slab.enter image description here

Electric field at a point between the plates due to charges $ + Q$and$ - Q$ on capacitor plates is ${E_0} = {\sigma \over {{\varepsilon _0}}} = {Q \over {A{\varepsilon _0}}}$ in a direction left to right.

Magnitude of induced charge:$${Q_p} = Q(1 - {1 \over K})$$ where K is dielectric constant. Or $$\overrightarrow E = {{\overrightarrow {{E_0}} } \over K}$$ where $\overrightarrow E $ is resultant electric field.For vacuum, there is no polarisation and hence $\overrightarrow E = \overrightarrow {{E_0}} $ and K=1.

  • $\begingroup$ What are q1, q2,q3,q4 going to be from my diagram ? $\endgroup$ – Shubham Nov 5 '15 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.