In the figure a nonconducting rod of length L = 8.38 cm has charge -q = -4.53 fC uniformly distributed along its length.

In the figure a nonconducting rod of length L = 8.38 cm has charge -q = -4.53 fC uniformly distributed along its length. (a) What is the linear charge density of the rod? What are the (b) magnitude and (c) direction (positive angle relative to the positive direction of the x axis) of the electric field produced at point P, at distance a = 13.7 cm from the rod? What is the electric field magnitude produced at distance a = 51 m by (d) the rod and (e) a particle of charge -q = -4.53 fC that replaces the rod?

The Correct Answer and Explanation is:

Let’s break down each part of the question:

Part (a) – Linear Charge Density of the Rod:

The linear charge density, λ\lambdaλ, is defined as the charge per unit length of the rod. It can be calculated using the formula:λ=qL\lambda = \frac{q}{L}λ=Lq​

Where:

• qqq is the total charge on the rod, which is −4.53 fC=−4.53×10−15 C-4.53 \, \text{fC} = -4.53 \times 10^{-15} \, \text{C}−4.53fC=−4.53×10−15C,
• LLL is the length of the rod, which is 8.38 cm=8.38×10−2 m8.38 \, \text{cm} = 8.38 \times 10^{-2} \, \text{m}8.38cm=8.38×10−2m.

Now substitute the values:λ=−4.53×10−158.38×10−2=−5.4×10−14 C/m\lambda = \frac{-4.53 \times 10^{-15}}{8.38 \times 10^{-2}} = -5.4 \times 10^{-14} \, \text{C/m}λ=8.38×10−2−4.53×10−15​=−5.4×10−14C/m

Thus, the linear charge density is λ=−5.4×10−14 C/m\lambda = -5.4 \times 10^{-14} \, \text{C/m}λ=−5.4×10−14C/m.

Part (b) – Magnitude of the Electric Field at Point P:

To find the magnitude of the electric field produced by the uniformly charged rod at a point located a distance aaa from it, we use the following formula for a uniformly charged line:E=14πϵ0⋅2λaE = \frac{1}{4\pi\epsilon_0} \cdot \frac{2\lambda}{a}E=4πϵ0​1​⋅a2λ​

Where:

• ϵ0\epsilon_0ϵ0​ is the permittivity of free space =8.85×10−12 C2/(N⋅m2)= 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)=8.85×10−12C2/(N⋅m2),
• a=13.7 cm=0.137 ma = 13.7 \, \text{cm} = 0.137 \, \text{m}a=13.7cm=0.137m,
• λ=−5.4×10−14 C/m\lambda = -5.4 \times 10^{-14} \, \text{C/m}λ=−5.4×10−14C/m.

Now, calculate the magnitude:E=14π(8.85×10−12)⋅2(−5.4×10−14)0.137E = \frac{1}{4\pi (8.85 \times 10^{-12})} \cdot \frac{2(-5.4 \times 10^{-14})}{0.137}E=4π(8.85×10−12)1​⋅0.1372(−5.4×10−14)​E=9.0×109⋅(−1.08×10−13)0.137E = \frac{9.0 \times 10^9 \cdot (-1.08 \times 10^{-13})}{0.137}E=0.1379.0×109⋅(−1.08×10−13)​E≈−7.1×103 N/CE \approx -7.1 \times 10^3 \, \text{N/C}E≈−7.1×103N/C

Thus, the magnitude of the electric field is approximately 7.1×103 N/C7.1 \times 10^3 \, \text{N/C}7.1×103N/C.

Part (c) – Direction of the Electric Field:

The electric field due to a negatively charged rod points toward the rod (since the charge is negative). If point PPP is on the positive x-axis, the electric field will point towards the rod, which means the electric field direction will be along the negative x-axis.

Part (d) – Electric Field Magnitude at Distance a=51 ma = 51 \, \text{m}a=51m:

For large distances (much greater than the length of the rod), we can approximate the rod as a point charge. The electric field from a point charge qqq at a distance rrr is given by Coulomb’s law:E=14πϵ0⋅∣q∣r2E = \frac{1}{4\pi\epsilon_0} \cdot \frac{|q|}{r^2}E=4πϵ0​1​⋅r2∣q∣​

Substitute the values:E=9.0×109⋅4.53×10−15512E = \frac{9.0 \times 10^9 \cdot 4.53 \times 10^{-15}}{51^2}E=5129.0×109⋅4.53×10−15​E=4.07×10−52601E = \frac{4.07 \times 10^{-5}}{2601}E=26014.07×10−5​E≈1.57×10−8 N/CE \approx 1.57 \times 10^{-8} \, \text{N/C}E≈1.57×10−8N/C

Thus, the electric field magnitude at a=51 ma = 51 \, \text{m}a=51m is approximately 1.57×10−8 N/C1.57 \times 10^{-8} \, \text{N/C}1.57×10−8N/C.

Part (e) – Electric Field Magnitude for a Point Charge −q-q−q:

Now, let’s calculate the electric field produced by a point charge of −q=−4.53 fC-q = -4.53 \, \text{fC}−q=−4.53fC at a distance of a=51 ma = 51 \, \text{m}a=51m:E=9.0×109⋅4.53×10−15512E = \frac{9.0 \times 10^9 \cdot 4.53 \times 10^{-15}}{51^2}E=5129.0×109⋅4.53×10−15​

This is exactly the same calculation as part (d), since the magnitude of the field depends only on the charge and the distance. Therefore, the electric field produced by a point charge is the same as for the rod at this distance:E≈1.57×10−8 N/CE \approx 1.57 \times 10^{-8} \, \text{N/C}E≈1.57×10−8N/C

Summary:

• (a) Linear charge density: λ=−5.4×10−14 C/m\lambda = -5.4 \times 10^{-14} \, \text{C/m}λ=−5.4×10−14C/m,
• (b) Electric field magnitude at a=13.7 cma = 13.7 \, \text{cm}a=13.7cm: 7.1×103 N/C7.1 \times 10^3 \, \text{N/C}7.1×103N/C,
• (c) Direction of electric field: along the negative x-axis,
• (d) Electric field magnitude at a=51 ma = 51 \, \text{m}a=51m from the rod: 1.57×10−8 N/C1.57 \times 10^{-8} \, \text{N/C}1.57×10−8N/C,
• (e) Electric field magnitude at a=51 ma = 51 \, \text{m}a=51m from a point charge: 1.57×10−8 N/C1.57 \times 10^{-8} \, \text{N/C}1.57×10−8N/C.