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A few pointers (since this is homework): A capacitor of capacitance $C$ charged to a voltage $V$ has charge $Q$ such that $$Q=CV$$ The positive side of capacitor 1 has charge $+Q_1$, the negative side has $-Q_1$ etc. Then the total charge (on the two capacitors) after connecting the capacitors depends on the polarity of the connection you made. If you ...


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I will give you hints how to do it. If you do it yourself, you will gain confidence. Charge will be conserved separately in the two plates you connect The charges will flow till potential difference across both capacitors is same. Note that equation $1$ will matter on which $2$ plates are connected. When you connect a resistor, the steady state will ...


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Your teacher is right. Although you missed a factor of $2$ in last line. The case you are saying is NOT equivalent to this one. How do you plan on filling dielectric $K_1$ on the lower half which is already filled with $K_2$? This makes no sense.


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It is essentially two capacitors in parallel. It will therefore have twice the capacitance than if everything else were held constant and one of the +Q plates were removed.


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http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-7-capacitance-and-field-energy/ Watch this video from 24:00 minutes till about 32:00 minutes (a total of about 8 minutes).


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Suppose first that $C_{5}$ is absent, that the voltage at the connection point of $C_{2}$ and $C_{3}$ is zero and that of the connection point of $% C_{1}$ and $C_{4}$ is $V(\omega )$. The impedance of a capacitor $C$ at the angular frequency $\omega $ is $$Z(C)=\frac{1}{i\omega C}.$$ Then \begin{eqnarray*} V(A) ...


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You need to assume how the current gets divided across the whole circuit. For example, take the current coming out of the main battery as $I$, then when it reaches the loop that you have selected, it breaks into $I_1$ and $I-I_1$. Do the same for all the loops, then apply Kirchoff's Rule. For capacitors, apply it exactly the same way you do it for batteries. ...


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In theory, provided that the surface area of the electrode remains constant and that the electrode material is conductive, the capacitance will not differ significantly when using different metals. This is just a consequence of the fact that for parallel plates, the capacitance only really depends on plate surface area. However, for modern electrochemical ...


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First of all that's $U$$=$$(1/2)$$C$$V^2$. But there is also another relation : $U$$=$$(1/2C)$$Q^2$. And I think your confusion will be gone if you realize when to use which relation. If you have a capacitor with a constant voltage source connected across it, then you use the first relation. But when a capacitor is already charged and is not part of a ...


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Once the battery is disconnected, the charge on the capacitor plates is stuck where it is and has no path to go anywhere else. Since the charge remains on the plates, there is an electric field between the plates. And because there's a electric field between the plates there must be a voltage difference between them. We know the voltage was equal to the ...


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If the plates are disconnected, the charge has nowhere to go. Rather U will have to change. What happens is the charged capacitor does work on the dielectric (pulling it in), resulting in a change in the energy stored in the capacitor.



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