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| | | ### Statement |
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| | − | $6.3.16.$ There is a charge $Q$ in the cavity of a metal ball of radius $R$. Find the charge induced by this charge on the surface of the cavity. Why will the charge be distributed with a constant density on the surface of the ball? What is the surface charge density of a sphere if its total charge is zero? Find the electric field strength outside the ball at a distance $L$ from its center if its total charge is $q$. Does this field depend on the location of the cavity in the ball? from its shape? |
| | + | $6.3.16.$ There is a charge $Q$ in the cavity of a metal ball of radius $R$. Find the charge induced by this charge on the surface of the cavity. Why will the charge be distributed with a constant density on the surface of the ball? What is the surface charge density of a sphere if its total charge is zero? Find the electric field strength outside the ball at a distance $L$ from its center if its total charge is $q$. Does this field depend on the location of the cavity in the ball? from its shape? |
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| | | ### Solution |
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| | − | The presence of a charge $Q$ in the cavity produces an electric field that penetrates the walls of the cavity, inducing a charge $Q'$. Since the ball is made of metal, the sphere is conducting, so charge on it resides on its surface, then the electric field inside of it is zero. Applying Gauss law, for a Gaussian sphere centered at the position of charge $Q$ and radius $r$, always inside the ball,\ |
| | − | $\int \vec{E}\cdot d\vec{S} = \frac{Q+Q'}{\varepsilon_0}$\ |
| | − | as $\vec{E} = \vec{0}$,\ |
| | − | $Q' = -Q$\ |
| | − | and the outer surface of ball is charged with $-Q' = +Q$.\ |
| | − | \ |
| | − | Let's take a Gaussian sphere centered at the ball's center with radius $R$, and consider the differential form of Gauss law,\ |
| | − | $\vec{E}\cdot d\vec{S} = \frac{dq}{\varepsilon_0}$\ |
| | − | $\sigma = \frac{dq}{dS} = \varepsilon_0 E(R)$\ |
| | − | but E(R) has a constant value independently on cavity position inside the ball. So, $\sigma$ is constant (**the distribution over the outer surface is uniform**).\ |
| | − | \ |
| | − | Applying Gauss law for a Gaussian sphere of radius $R$ centered at the ball (total charge = $Q'-Q'$ = 0), |
| | − | $\int\vec{E}\cdot d\vec{S} = \frac{-Q'+Q'+Q}{\varepsilon_0} = \frac{Q}{\varepsilon_0}$ (1)\ |
| | − | $\sigma = \frac{dq}{dS} = \varepsilon_0 E(R)$ (2)\ |
| | − | From (1),\ |
| | − | $E(R) = \frac{Q}{4\pi\varepsilon_0 R^2}$ (3)\ |
| | − | Putting (3) into (2),\ |
| | − | $\sigma = \frac{Q}{4\pi R^2}$\ |
| | − | \ |
| | − | Applying Gauss law for a sphere centered at ball and radius $r > R$ and evaluating at $r=L$,\ |
| | − | $\int \vec{E}\cdot d\vec{S} = \frac{q+Q}{\varepsilon_0}$\ |
| | − | $E(L) = \frac{q+Q}{4\pi\varepsilon_0 L^2}$\ |
| | − | The value of E(L) **doesn't depend on the cavity's position nor size of sphere, but the enclosed charge**. |
| | + | <b>1. Induced charge on the cavity surface:</b> |
| | + | Inside a conductor (within the bulk of the metal), the electrostatic field is always zero ($E = 0$). Let's enclose the cavity with a closed Gaussian surface passing entirely within the metal of the ball. According to Gauss's Law, the total enclosed charge must be zero. Therefore, the charge $Q$ in the cavity induces a charge on the inner surface of the cavity equal to: |
| | + | $$Q_{\text{inner}} = -Q$$ |
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| | − | #### Answer |
| | + | <b>2. Charge distribution on the outer surface:</b> |
| | + | Since the electric field inside the metal is zero, the outer surface of the ball is completely shielded from whatever happens inside the cavity. For an outside observer, the metal ball is a perfectly spherical equipotential surface. The absence of external fields forces the induced charge on the outer surface to distribute itself completely uniformly due to spherical symmetry. |
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| | − | $Q' = -Q$\ |
| | − | Charge distribution obver the outer surface is uniform\ |
| | − | $\sigma = \frac{Q}{4\pi R^2}$\ |
| | − | $E(L) = \frac{q+Q}{4\pi\varepsilon_0 L^2}$\ |
| | − | NO, NO |
| | + | <b>3. Surface density if the total charge of the ball is zero:</b> |
| | + | If the net charge of the metal ball itself is zero, the appearance of a $-Q$ charge on the inner cavity surface means that a compensating charge of $+Q$ must appear on the outer surface (law of conservation of charge). |
| | + | Since it is distributed uniformly over a sphere of radius $R$, its surface charge density is: |
| | + | $$\sigma = \frac{+Q}{S} = \frac{Q}{4\pi R^2}$$ |
| | + | |
| | + | <b>4. Electric field outside at distance $L$ if the ball's charge is $q$:</b> |
| | + | If the metal ball has its own net charge $q$, then according to the conservation of charge, the total charge accumulated on its outer surface will be $(Q + q)$. |
| | + | Applying Gauss's Law for a spherical surface of radius $L > R$, we find that the field outside the ball is equivalent to the field of a point charge located at the center: |
| | + | $$E(L) = \frac{1}{4\pi\varepsilon_0} \frac{Q + q}{L^2}$$ |
| | + | |
| | + | <b>5. Dependence on the cavity:</b> |
| | + | The electric field outside the ball <b>does not depend</b> on the location of the cavity, nor on its shape. The metal layer completely "forgets" the geometry of the interior, acting as a perfect shield. The external field is determined solely by the shape of the outer surface of the ball and the total enclosed charge. |
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| | + | #### Answer |
| | + | $Q_{\text{inner}} = -Q$; $\sigma = \frac{Q}{4\pi R^2}$; $E(L) = \frac{1}{4\pi\varepsilon_0} \frac{Q + q}{L^2}$; the external field does not depend on the location and shape of the cavity. |
