An polaroid long cylindrical object with radius R has a charge distribution that depends upon distance r from the axis like this : `rho = ar + br^(2)(r le R`, a and b are non zero constant, `rho` is volume charge density). If electric field outside the cylinder is zero than value of `(a)/(b)` is :

Options

A. `3R//4`
B. `-3R//4`
C. `-4R//3`
D. `4 R//3`

Explanation

To solve the problem, we need to determine the ratio \( \frac{a}{b} \) given the charge density distribution \( \rho = ar + br^2 \) for a long cylindrical object with radius \( R \), under the condition that the electric field outside the cylinder is zero.

### Step-by-Step Solution:

1. **Understand the Problem**: We know that the electric field outside the cylinder is zero. According to Gauss's law, this implies that the total charge enclosed within a Gaussian surface (which is outside the cylinder) must be zero.

2. **Set Up the Charge Density**: The volume charge density is given by:
\[ \rho(r) = ar + br^2 \quad (0 \leq r \leq R) \]

3. **Calculate Total Charge Inside the Cylinder**: The total charge \( Q \) inside the cylinder can be found by integrating the charge density over the volume of the cylinder:
\[ Q = \int_0^R \rho(r) \cdot dV \]
where \( dV = 2\pi r \, dr \) (the differential volume element for a cylindrical shell).

4. **Substituting for \( dV \)**:
\[ Q = \int_0^R (ar + br^2) \cdot (2\pi r) \, dr \]
\[ Q = 2\pi \int_0^R (ar^2 + br^3) \, dr \]

5. **Evaluate the Integral**:
\[ Q = 2\pi \left[ \frac{a r^3}{3} + \frac{b r^4}{4} \right]_0^R \]
\[ Q = 2\pi \left( \frac{a R^3}{3} + \frac{b R^4}{4} \right) \]

6. **Set Total Charge to Zero**: Since the electric field outside the cylinder is zero, we set \( Q = 0 \):
\[ \frac{a R^3}{3} + \frac{b R^4}{4} = 0 \]

7. **Rearranging the Equation**:
\[ \frac{a R^3}{3} = -\frac{b R^4}{4} \]

8. **Finding the Ratio \( \frac{a}{b} \)**:
\[ \frac{a}{b} = -\frac{3R}{4} \]

### Final Answer:
The value of \( \frac{a}{b} \) is:
\[ \frac{a}{b} = -\frac{3R}{4} \]