Question

A 10-cm-long thin glass rod uniformly charged to 14.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 14.0 nC are placed side by side, 4.50 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?

Answer 1

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

Answer 2

The electric fields are:

(i) At a point 1cm from the glass rod, E = 1.36 * 10⁶N/C

(ii) At a point 2cm from the glass rod, E = 5.17 * 10⁵N/C

(iii) At a point 3cm from the glass rod, E = 6.993 * 10⁵N/C

From the question we get that:

Charge of glass rod, Q = 14nC = 14 * 10⁻⁹ C

Charge of plastic rod, q = 14nC = 14 * 10⁻⁹ C

Distance between the rods = 4.5cm = 0.045

(i) Electric field strength at a point 1.0cm from the glass rod

will be the sum of electric field strength due to both the rods.

Electric field is given by:

E = kQ/r²

here, Q is the charge,

r is the distance, and

k = 9 × 10⁹ Nm²/C²

E = E₁ + E₂

where, E₁ = electric field strength due to glass rod

E₂ = electric field strength due to plastic rod

$E_1 = kQ/0.01^2\\ \\ E_2 = kq/(0.045 - 0.01)^2 = kq/(0.035)^2\\ \\ E = kQ/0.01^2 + kq/(0.035)^2 = k(Q/0001 + q/0.001225)\\ \\ E = 9 * 10^9 [(14* 10^{-9} / 0.001) + (14 * 10^{-9})/0.001225]\\ \\ E = 9 *10^9[(14 * 10^{-5}) + (1.143 * 10^{-5})]\\ \\ E = 9*10^9* 15.143* 10^{-5}\\ \\ E = 1.36*10^6N/C$

(ii) Electric field strength at a point 2.0cm from the glass rod

$E = E_1 + E_2\\ \\ E_1 = kQ/0.02^2\\ \\ E_2 = kq/(0.045 - 0.02)^2 = kq/(0.025)^2\\ \\ E = kQ/0.02^2 + kq/(0.025)^2 = k(Q/0.0004 + q/0.000625)\\ \\ E = 9 * 10^9 [(14 * 10^{-9} / 0.0004) + (14 * 10^{-9})/0.000625]\\ \\ E = 9 * 10^9[(3.5 * 10^{-5}) + (2.24 * 10^{-5})]\\ \\ E = 9 * 10^9 * 5.74 * 10^{-5}\\ \\ E = 5.17 * 10^5N/C$

(III) Electric field strength at a point 3.0cm from the glass rod

$E = E_1 + E_2\\ \\ E_1 = kQ/0.03^2\\ \\ E_2 = kq/(0.045 - 0.03)^2 = kq/(0.015)^2\\ \\ E = kQ/0.03^2 + kq/(0.015)^2 = k(Q/0009 + q/0.000225)\\\\ E = 9 * 10^9 [(14 * 10^{-9} / 0.009) + (14 * 10^{-9})/0.000225]\\ \\ E = 9 * 10^9[( 1.55 * 10^{-5}) + (6.22 * 10^{-5})]\\ \\E = 9 * 10^9 * 7.77 * 10^{-5}\\\\ E = 6.993 * 10^5N/C$

Hence, electric fields for all the situations are as discussed above.

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