Answer:
A. At a point 1cm from the glass rod, E = 1.36 * 10⁶N/C
B. At a point 2cm from the glass rod, E = 5.17 * 10⁵N/C
C. At a point 3cm from the glass rod, E = 6.993 * 10⁵N/C
Explanation:
Parameters given:
Charge of glass rod, Q = 14nC = 14 * 10⁻⁹ C
Charge of plastic rod, q = 14nC = 14 * 10⁻⁹ C
Distance between both rods = 4.5cm = 0.045
A. Electric field strength at a point 1.0cm (0.01m) from the glass rod is the sum of electric field strength due to both rods i.e.
E = E₁ + E₂
Where
E₁ = electric field strength due to glass rod
E₂ = electric field strength due to plastic rod
E₁ = kQ/0.01²
E₂ = kq/(0.045 - 0.01)² = kq/(0.035)²
E = kQ/0.01² + kq/(0.035)² = k(Q/0001 + q/0.001225)
E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.001) + (14 * 10⁻⁹)/0.001225]
E = 9 * 10⁹[(14 * 10⁻⁵) + (1.143 * 10⁻⁵)]
E = 9 * 10⁹ * 15.143* 10⁻⁵
E = 1.36 * 10⁶N/C
B. Electric field strength at a point 2.0cm (0.02m) from the glass rod is the sum of electric field strength due to both rods i.e.
E = E₁ + E₂
E₁ = kQ/0.02²
E₂ = kq/(0.045 - 0.02)² = kq/(0.025)²
E = kQ/0.02² + kq/(0.025)² = k(Q/0.0004 + q/0.000625)
E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.0004) + (14 * 10⁻⁹)/0.000625]
E = 9 * 10⁹[(3.5 * 10⁻⁵) + (2.24 * 10⁻⁵)]
E = 9 * 10⁹ * 5.74 * 10⁻⁵
E = 5.17 * 10⁵N/C
C. Electric field strength at a point 3.0cm (0.03m) from the glass rod is the sum of electric field strength due to both rods i.e.
E = E₁ + E₂
E₁ = kQ/0.03²
E₂ = kq/(0.045 - 0.03)² = kq/(0.015)²
E = kQ/0.03² + kq/(0.015)² = k(Q/0009 + q/0.000225)
E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.009) + (14 * 10⁻⁹)/0.000225]
E = 9 * 10⁹[( 1.55 * 10⁻⁵) + (6.22 * 10⁻⁵)]
E = 9 * 10⁹ * 7.77 * 10⁻⁵
E = 6.993 * 10⁵N/C
