Understanding Charge Distribution and Electrostatic Forces
This problem delves into the principles of electrostatics, specifically focusing on calculating the forces within a system comprising a sphere with a non-uniform charge distribution and two discrete point charges. The core task is to determine the distance '$a$' at which these point charges experience no net force. This involves applying concepts like Gauss's Law to find the electric field of the charged sphere and understanding Coulomb's Law for the forces between point charges.
Calculating the Constant C from Charge Density
We are given a sphere of radius R with a charge density described by the function $\rho(r) = Cr^2$, where r denotes the distance from the center of the sphere. The total charge contained within this sphere is specified as $2Q$. To proceed, we first need to determine the value of the constant C in the charge density expression.
The total charge ($Q_{total}$) can be found by integrating the charge density over the entire volume of the sphere. In spherical coordinates, the differential volume element is $dV = 4\pi r^2 dr$. Therefore, the total charge is:
$ Q_{total} = \int_{0}^{R} \rho(r) dV = \int_{0}^{R} (Cr^2)(4\pi r^2 dr) $Combining the terms, we get:
$ Q_{total} = 4\pi C \int_{0}^{R} r^4 dr $Evaluating the integral:
$ Q_{total} = 4\pi C \left[ \frac{r^5}{5} \right]_{0}^{R} = \frac{4\pi C R^5}{5} $Given that the total charge $Q_{total}$ is $2Q$, we set up the equation:
$ 2Q = \frac{4\pi C R^5}{5} $Solving this equation for C yields:
$ C = \frac{5 \times 2Q}{4\pi R^5} = \frac{10Q}{4\pi R^5} = \frac{5Q}{2\pi R^5} $Determining the Electric Field of the Sphere
To calculate the force exerted by the charged sphere on the point charges $A$ and $B$, we must first find the electric field generated by the sphere. Due to the spherical symmetry of the charge distribution (depending only on r), the electric field $\vec{E}$ will be purely radial. We can utilize Gauss's Law, which states $\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$, to find the magnitude of the electric field, $E(r)$.
Consider a spherical Gaussian surface with radius r, concentric with the charged sphere. The electric field is radial ($\vec{E} = E(r) \hat{r}$) and the area element is also radial ($d\vec{A} = dA \hat{r}$). Therefore, the flux integral simplifies to $E(r) \times (\text{Surface Area})$:
$ \oint \vec{E} \cdot d\vec{A} = E(r) (4\pi r^2) $Now, we need the charge enclosed ($q_{enc}$) within this Gaussian surface. Assuming the point charges are placed at a distance '$a$' such that $a \le R$, the enclosed charge is calculated by integrating the density from the center up to radius r:
$ q_{enc}(r) = \int_{0}^{r} \rho(r') dV' = \int_{0}^{r} (Cr'^2)(4\pi r'^2 dr') $ $ q_{enc}(r) = 4\pi C \int_{0}^{r} r'^4 dr' = 4\pi C \frac{r^5}{5} $Applying Gauss's Law ($E(r) (4\pi r^2) = \frac{q_{enc}}{\epsilon_0}$):
$ E(r) (4\pi r^2) = \frac{4\pi C r^5}{5 \epsilon_0} $Solving for the electric field magnitude $E(r)$:
$ E(r) = \frac{C r^3}{5 \epsilon_0} $Substituting the expression for C we found earlier ($C = \frac{5Q}{2\pi R^5}$):
$ E(r) = \frac{1}{5 \epsilon_0} \left( \frac{5Q}{2\pi R^5} \right) r^3 = \frac{Q r^3}{2\pi \epsilon_0 R^5} $This formula gives the electric field strength at a distance r from the center for $r \le R$. Let $E_{sphere}(a)$ denote the electric field at the location of charges A and B:
$ E_{sphere}(a) = \frac{Q a^3}{2\pi \epsilon_0 R^5} $Analyzing Forces on Charges A and B
We have two point charges, $A$ and $B$, each carrying a charge of $-Q$. They are positioned diametrically opposite to each other at an equal distance '$a$' from the center of the sphere. The crucial condition is that these charges experience no net force. This means the vector sum of all forces acting on charge $A$ must be zero, and similarly for charge $B$. Due to symmetry, if the net force on $A$ is zero, the net force on $B$ will also be zero.
Let's focus on charge $A$. The forces acting on it are:
- Force from the sphere ($\vec{F}_{sphere \to A}$): This force arises from the electric field generated by the sphere's charge distribution at the location of $A$.
- Force from charge B ($\vec{F}_{B \to A}$): This is the electrostatic Coulomb force between charge $A$ and charge $B$.
The net force on $A$ is $\vec{F}_{net, A} = \vec{F}_{sphere \to A} + \vec{F}_{B \to A}$. The problem states $\vec{F}_{net, A} = \vec{0}$.
Force Exerted by the Sphere
The force on charge $A$ (with charge $-Q$) due to the sphere's electric field $\vec{E}_{sphere}(a)$ is given by:
$ \vec{F}_{sphere \to A} = (-Q) \vec{E}_{sphere}(a) $Since $\vec{E}_{sphere}(a)$ is radial, we can write $\vec{E}_{sphere}(a) = E_{sphere}(a) \hat{r}_A$, where $\hat{r}_A$ is the unit vector pointing radially outward from the center towards $A$. Thus:
$ \vec{F}_{sphere \to A} = -Q E_{sphere}(a) \hat{r}_A $The negative sign indicates that the force exerted by the sphere on the negative charge $-Q$ is opposite to the direction of the sphere's electric field.
Force Between Charges A and B
Charges $A$ and $B$ are located at position vectors $\vec{r}_A$ and $\vec{r}_B$, respectively. They are diametrically opposite, meaning $\vec{r}_B = -\vec{r}_A$, and both are at distance $a$ from the center, so $|\vec{r}_A| = |\vec{r}_B| = a$. The vector separation between them is $\vec{r}_A - \vec{r}_B = \vec{r}_A - (-\vec{r}_A) = 2\vec{r}_A$. The distance between them is $|\vec{r}_A - \vec{r}_B| = |2\vec{r}_A| = 2a$. Using Coulomb's Law, the force exerted by charge $B$ ($-Q$) on charge $A$ ($-Q$) is:
$ \vec{F}_{B \to A} = k \frac{(-Q)(-Q)}{|\vec{r}_A - \vec{r}_B|^2} \frac{\vec{r}_A - \vec{r}_B}{|\vec{r}_A - \vec{r}_B|} $Substituting the distance and vector separation:
$ \vec{F}_{B \to A} = k \frac{Q^2}{(2a)^2} \frac{2\vec{r}_A}{2a} = k \frac{Q^2}{4a^2} \frac{2\vec{r}_A}{2a} $ $ \vec{F}_{B \to A} = k \frac{Q^2}{4a^3} \vec{r}_A $Expressing this force in terms of the radial unit vector $\hat{r}_A = \vec{r}_A / a$:
$ \vec{F}_{B \to A} = k \frac{Q^2}{4a^3} (a \hat{r}_A) = k \frac{Q^2}{4a^2} \hat{r}_A $This force is repulsive (since both charges are negative) and directed radially outward along the line connecting the center to charge $A$.
Applying the Zero Force Condition
For charge $A$ to experience zero net force, the vector sum of the forces acting on it must be zero:
$ \vec{F}_{net, A} = \vec{F}_{sphere \to A} + \vec{F}_{B \to A} = \vec{0} $Substituting the expressions for the forces:
$ -Q E_{sphere}(a) \hat{r}_A + k \frac{Q^2}{4a^2} \hat{r}_A = \vec{0} $Since $\hat{r}_A$ is a non-zero unit vector, the scalar components must sum to zero:
$ -Q E_{sphere}(a) + k \frac{Q^2}{4a^2} = 0 $Rearranging this equation gives the condition on the electric field strength:
$ Q E_{sphere}(a) = k \frac{Q^2}{4a^2} $ $ E_{sphere}(a) = k \frac{Q}{4a^2} $Now, we equate this required field strength with the field produced by the sphere at distance '$a$', using the expression derived earlier ($E_{sphere}(a) = \frac{Q a^3}{2\pi \epsilon_0 R^5}$), assuming $a \le R$:
$ \frac{Q a^3}{2\pi \epsilon_0 R^5} = k \frac{Q}{4a^2} $Using the relation $k = \frac{1}{4\pi \epsilon_0}$:
$ \frac{Q a^3}{2\pi \epsilon_0 R^5} = \frac{1}{4\pi \epsilon_0} \frac{Q}{4a^2} $We can cancel out common terms ($Q$, $\pi$, and $\epsilon_0$) from both sides:
$ \frac{a^3}{2 R^5} = \frac{1}{16 a^2} $Now, we solve for $a$. Multiply both sides by $16a^2 R^5$:
$ 16 a^2 \cdot a^3 = 2 R^5 $ $ 16 a^5 = 2 R^5 $Isolating $a^5$:
$ a^5 = \frac{2}{16} R^5 = \frac{1}{8} R^5 $Taking the fifth root of both sides to find $a$:
$ a = \left( \frac{1}{8} \right)^{1/5} R $This can be written as:
$ a = 8^{-1/5} R $Verifying the Assumption $a \le R$
Let's quickly check what happens if we assumed $a > R$. In that case, the Gaussian surface of radius $a$ would enclose the entire charge of the sphere, $2Q$. The electric field outside the sphere ($r > R$) would be:
$ E_{sphere}(a) = \frac{1}{4\pi \epsilon_0} \frac{2Q}{a^2} = k \frac{2Q}{a^2} $Applying the zero net force condition $E_{sphere}(a) = k \frac{Q}{4a^2}$:
$ k \frac{2Q}{a^2} = k \frac{Q}{4a^2} $This simplifies to $2 = \frac{1}{4}$, which is impossible. Therefore, our initial assumption that $a \le R$ must be correct.
Final Conclusion
Based on the analysis of the forces and applying the condition for zero net force, the distance '$a$' from the center at which charges $A$ and $B$ experience no force is determined to be $a = 8^{-1/5}R$.
Important Questions from Electric Fields and Gauss' Law
A positive charge +q is placed at the centre of a hollow metallic sphere of inner radius a and outer radius b. the electric field at a distance r from the centre is denoted by E. In this regards, which one of the following statement is correct?
If a free electron moves through a potential difference of 1 kV, then the energy gained by the electron is given by
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$\left[ {{\rm{Take\;}}\frac{1}{{4{\rm{\pi }}{\epsilon_0}}} = 9 \times {{10}^9}{\rm{N}}{{\rm{m}}^2}{{\rm{C}}^{ - 2}}} \right]$ The expression for torque '\(\vec{\tau}\)' experienced by an electric dipole of dipole moment '\(\vec{P}\)' in an external uniform electric field '\(\vec{E}\)' is given by :
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