Electric field inside capacitor with dielectric, spotting the mistake in solution

1 vote
63 views Attempt:

$V_1 + V_2 = 440$
$Q_1= Q_2 \implies k_1V_1C_1 = k_2V_2C_2$
$\implies V_2 = \dfrac{k_1C_1V_1}{k_2C_2}= \dfrac{9V_1}{35}$
$\implies V_1 = 350 V$
$E_1 = \dfrac{V_1}{d_1 k_1} = \dfrac{350}{0.07 \times 3}= \dfrac{5000}{3}$

But answers are: $E_1 = 5\times 10^4 V/m, E_2 = 3\times 10^4 V/m$

Note: I am using $E_1= \dfrac{V_1}{d_1k_1}$ because electric field in the presence of a dielectric is given by $\dfrac{E_o}{k}$ right? If I am misusing it please let me know. Also, what's the correct way to use this formula in this problem?

edited Jul 18, 2018

1 vote

Electric field inside the capacitor is $V/d$. If $V, d$ remain constant then inserting dielectric between the plates makes no difference to total electric field inside the dielectric $E=V/d$.

The effect the dielectric has is to increase the free charge on the plates.

$$V_1 + V_2 = V_0 =440V$$
$$Q_1= Q_2 \implies k_2C_2 V_2= k_1 C_1 V_1=k_1 C_1 (V_0-V_2)$$
$$V_2=\dfrac{k_1 C_1}{k_1C_1+k_2C_2} V_0=\dfrac{k_1d_2}{k_1d_2+k_2d_1}V_0$$
$$V_1=\dfrac{k_2 d_1}{k_1 d_2+k_2 d_1}V_0$$
because $\dfrac{C_2}{C_1}=\dfrac{d_1}{d_2}$. The electric field in each half of the capacitor is $$E_2=\dfrac{V_2}{d_2}=\dfrac{k_1}{k_1d_2+k_2 d_1} V_0$$ $$E_1=\dfrac{V_1}{d_1}=\dfrac{k_2}{k_1d_2+k_2d_1}V_0$$

The ratio of stored energies is $$\frac{C_1 V_1^2}{C_2 V_2^2}=\frac{d_2}{d_1}(\frac{k_2 d_1}{k_1d_2})^2=\dfrac{d_1}{d_2}(\dfrac{k_2}{k_1})^2$$

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answered Jul 18, 2018 by (28,448 points)
selected Jul 18, 2018
thanks a lot :)! I didn't know that electric field remains constant when dielectric is inserted with battery connected, now I know it!