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Question: A point charge \(50\mu C\) is located in the \(XY\) plane at the point of position vector \({\vec r_...

A point charge 50μC50\mu C is located in the XYXY plane at the point of position vector r⃗0=2i^+3j^{\vec r_0} = 2\hat i + 3\hat j. What is the electric field at the point vector r⃗0=8i^−5j^{\vec r_0} = 8\hat i - 5\hat j.
(A) 1200 V/m1200\,V/m
(B) 0.04 V/m0.04\,V/m
(C) 900 V/m900\,V/m
(D) 4500 V/m4500\,V/m

Explanation

Solution

From the given two vectors, calculate the distance between the charge and the point where the electric field is calculated by subtracting those vectors. Use the formula of the distance calculated to substitute in the formula and calculate the value of the electric field.

Useful formula:
The electric field due to a point charge is given by
E=q4π∈0r2E = \dfrac{q}{{4\pi { \in _0}{r^2}}}
Where EE is the electric field due to a point charge, qq is the charge, ∈0{ \in _0} is the permittivity of the free space and rr is the distance of the point from the charge.

Complete step by step solution:
First, the distance of the point from the charge is to be calculated. Taking the two vectors given for calculating the value of rr.

r⃗=(8−2)i^−(5+3)j^{\vec r_{}} = \left( {8 - 2} \right)\hat i - \left( {5 + 3} \right)\hat j
Subtracting these two vectors gives the distance between them.
r⃗=6i^−8j^\vec r = 6\hat i - 8\hat j
In order to remove the vectors on both sides, take the modulus on both sides.
∣r⃗∣=∣6i^−8j^∣\left| {\vec r} \right| = \left| {6\hat i - 8\hat j} \right|
r=(62+82)r = \sqrt {\left( {{6^2} + {8^2}} \right)}
By performing the simplification,
r=10 mr = 10\,m
By using the formula of the electric field due to the point charge and substituting the required parameters there.
E=50×10−64π×8.854×10−12×102E = \dfrac{{50 \times {{10}^{ - 6}}}}{{4\pi \times 8.854 \times {{10}^{ - 12}} \times {{10}^2}}}
By performing the basic arithmetic operation,
E=504×3.14×8.854×10−4E = \dfrac{{50}}{{4 \times 3.14 \times 8.854 \times {{10}^{ - 4}}}}
E=4496.15 Vm−1E = 4496.15\,V{m^{ - 1}}
E≈4500 Vm−1E \approx 4500\,V{m^{ - 1}}.
Hence the electric field due to the point charge is 4500 Vm−14500\,V{m^{ - 1}}

Thus the option (D) is correct.

Note: The charge is given as 50μC50\mu C. It must be converted into coulomb to calculate in the formula as 50×10−6  C50 \times {10^{ - 6\,}}\,C. This is because micro denotes the value of 10−6{10^{ - 6}}. Remember that in order to calculate the distance between two vectors, the vector of the charge must be subtracted from the vector of the point.