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<meta name="description" content="a) The surface densities of charges in two planes are \pm \sigma. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act? b) The intensity of electric field between two parallel planes is null, out of them is E. Find the surface density of charge in the planes. What electric pressure acts over the planes? c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.">
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<meta property="og:description" content="a) The surface densities of charges in two planes are \pm \sigma. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act? b) The intensity of electric field between two parallel planes is null, out of them is E. Find the surface density of charge in the planes. What electric pressure acts over the planes? c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.">
<title>a) The surface densities of charges in two planes are \pm\sigma. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act? b) The intensity of electric field between two parallel planes is null, out of them is E. Find the surface density of charge in the planes. What electric pressure acts over the planes? c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.</title>
a) The surface densities of charges in two planes are $\pm\sigma$. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act?
</p>
<p class="TxtSolutions">
b) The intensity of electric field between two parallel planes is null, out of them is $E$. Find the surface density of charge in the planes. What electric pressure acts over the planes?
</p>
<p class="TxtSolutions">
c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.
</p>
<h3>Solution</h3>
<p>
a) The absolute value of the electric field created by each plate is $E = \frac{\sigma}{2\varepsilon_0}$, between them is $2E = \frac{\sigma}{\varepsilon_0}$ (superposition). The force that acts over a differential of charge $dq$ is
According to superposition principle between plates, $\vec{E_1}+\vec{E_2} = \vec{0}$, hence, $\vec{E_1}=-\vec{E_2}$ or $E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$. So, plates must have charges of same sign (+) or (-). The outside field is
$$\vec{E} = \vec{E_1} + \vec{E_2}$$
$E = \frac{\sigma}{\varepsilon_0}$
$$\boxed{\sigma = \varepsilon_0 E} \;(1)$$
The force that acts over an element of charge $dq$ in a plate is $dF_i = E_i~dq$ (i =1,2), but $E_i = \frac{E}{2}$ so,
Superposition of electric fields must be zero outside of capacitor, so $\vec{E_1} = -\vec{E_2}$, so plates must have opposed signs. Inside is valid that $\vec{E_1} = \vec{E_2}$ so,
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
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<meta name="description" content="The largest dataset of solutions of 'Savchenko. Problems in Physics'. Savchenko’s Problems in General Physics is widely used to prepare for olympiads and it is a useful tool to
<meta name="description" content="a) The surface densities of charges in two planes are \pm \sigma. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act? b) The intensity of electric field between two parallel planes is null, out of them is E. Find the surface density of charge in the planes. What electric pressure acts over the planes? c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.">
master and sharpen your skills and techniques in comptetitive problem solving. Some of these problems were a source
of inspiration for Jaan Kalda’s handouts and to some NBPhO problems. You may find problems from old IPhO
<meta property="og:title" content="a) The surface densities of charges in two planes are \pm \sigma. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act? b) The intensity of electric field between two parallel planes is null, out of them is E. Find the surface density of charge in the planes. What electric pressure acts over the planes? c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.">
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<meta property="og:description" content="A website with solutions to physics problems from Savchenko Textbook">
<meta property="og:description" content="a) The surface densities of charges in two planes are \pm \sigma. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act? b) The intensity of electric field between two parallel planes is null, out of them is E. Find the surface density of charge in the planes. What electric pressure acts over the planes? c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.">
<title>a) The surface densities of charges in two planes are \pm\sigma. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act? b) The intensity of electric field between two parallel planes is null, out of them is E. Find the surface density of charge in the planes. What electric pressure acts over the planes? c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.</title>
a) The surface densities of charges in two planes are $\pm\sigma$. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act?
a) The surface densities of charges in two planes are $\pm\sigma$. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act?
</p>
</p>
<p class="TxtSolutions">
<p class="TxtSolutions">
b) The intensity of electric field between two parallel planes is null, out of them is $E$. Find the surface density of charge in the planes. What electric pressure acts over the planes?
b) The intensity of electric field between two parallel planes is null, out of them is $E$. Find the surface density of charge in the planes. What electric pressure acts over the planes?
</p>
</p>
<p class="TxtSolutions">
<p class="TxtSolutions">
c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.
c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.
</p>
</p>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
a) The absolute value of the electric field created by each plate is $E = \frac{\sigma}{2\varepsilon_0}$, between them is $2E = \frac{\sigma}{\varepsilon_0}$ (superposition). The force that acts over a differential of charge $dq$ is
a) The absolute value of the electric field created by each plate is $E = \frac{\sigma}{2\varepsilon_0}$, between them is $2E = \frac{\sigma}{\varepsilon_0}$ (superposition). The force that acts over a differential of charge $dq$ is
According to superposition principle between plates, $\vec{E_1}+\vec{E_2} = \vec{0}$, hence, $\vec{E_1}=-\vec{E_2}$ or $E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$. So, plates must have charges of same sign (+) or (-). The outside field is
According to superposition principle between plates, $\vec{E_1}+\vec{E_2} = \vec{0}$, hence, $\vec{E_1}=-\vec{E_2}$ or $E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$. So, plates must have charges of same sign (+) or (-). The outside field is
$$\vec{E} = \vec{E_1} + \vec{E_2}$$
$$\vec{E} = \vec{E_1} + \vec{E_2}$$
$E = \frac{\sigma}{\varepsilon_0}$
$E = \frac{\sigma}{\varepsilon_0}$
$$\boxed{\sigma = \varepsilon_0 E} \;(1)$$
$$\boxed{\sigma = \varepsilon_0 E} \;(1)$$
The force that acts over an element of charge $dq$ in a plate is $dF_i = E_i~dq$ (i =1,2), but $E_i = \frac{E}{2}$ so,
The force that acts over an element of charge $dq$ in a plate is $dF_i = E_i~dq$ (i =1,2), but $E_i = \frac{E}{2}$ so,
Superposition of electric fields must be zero outside of capacitor, so $\vec{E_1} = -\vec{E_2}$, so plates must have opposed signs. Inside is valid that $\vec{E_1} = \vec{E_2}$ so,
Superposition of electric fields must be zero outside of capacitor, so $\vec{E_1} = -\vec{E_2}$, so plates must have opposed signs. Inside is valid that $\vec{E_1} = \vec{E_2}$ so,
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
