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A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$.

A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$.

I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's lawCoulomb's law, won't the force be defined as
$$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2~ ?$$

A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$.

I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's law, won't the force be defined as
$$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2~ ?$$

A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$.

I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's law, won't the force be defined as
$$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2~ ?$$

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A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$.

A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$. I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's law, won't the force be defined as
$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2$ ?$$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2~ ?$$

A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$. I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's law, won't the force be defined as
$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2$ ?

A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$.

I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's law, won't the force be defined as
$$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2~ ?$$

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Saprativ Ray
Saprativ Ray
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A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$. I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's law, won't the force be defined as
$F=\frac{1}{4\pi\epsilon_0} \frac{CV\times CV}{d^2} = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2}$$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2$ ?

A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$. I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's law, won't the force be defined as
$F=\frac{1}{4\pi\epsilon_0} \frac{CV\times CV}{d^2} = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2}$ ?

A textbook question requires me to calculate the force of attraction between plates of a parallel-plate capacitor. The answer provided is $\frac{1}{2}QE$. I am not entirely sure how they arrived at it. The charge on each plate will be $Q=CV$ so from Coulomb's law, won't the force be defined as
$F=\frac{1}{4\pi\epsilon_0} (\frac{CV}{d})^2 = \frac{1}{4\pi\epsilon_0} (\frac{Q}{d})^2$ ?

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Saprativ Ray
Saprativ Ray
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