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

1 vote


$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?

asked Jul 18, 2018 in Physics Problems by Reststack (422 points)
edited Jul 18, 2018 by Reststack

1 Answer

1 vote
Best answer

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.

See :
Electric field intensity in a dielectric inside a capacitor
Electric field in a capacitor with a dielectric with variable permittivity
Filling a charged capacitor with dielectric material
What happens when a dielectric is inserted in a capacitor connected to a battery?

$$V_1 + V_2 = V_0 =440V $$
$$Q_1= Q_2 \implies k_2C_2 V_2= k_1 C_1 V_1=k_1 C_1 (V_0-V_2)$$
$$V_2=\dfrac{k_1 C_1}{k_1C_1+k_2C_2} V_0=\dfrac{k_1d_2}{k_1d_2+k_2d_1}V_0$$
$$V_1=\dfrac{k_2 d_1}{k_1 d_2+k_2 d_1}V_0$$
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 $$\frac{C_1 V_1^2}{C_2 V_2^2}=\frac{d_2}{d_1}(\frac{k_2 d_1}{k_1d_2})^2=\dfrac{d_1}{d_2}(\dfrac{k_2}{k_1})^2$$


answered Jul 18, 2018 by sammy gerbil (28,896 points)
selected Jul 18, 2018 by Reststack
thanks a lot :)! I didn't know that electric field remains constant when dielectric is inserted with battery connected, now I know it!