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Question: A charge \(q=\,-2.0\mu C\) is placed at origin. Find the electric field at (3m, 4m, 0)....

A charge q=2.0μCq=\,-2.0\mu C is placed at origin. Find the electric field at (3m, 4m, 0).

Explanation

Solution

This is an electrostatics question in which we need to find an electric field and for calculating this we need to know the formula of the electric field. From the given data we need to find vector r and resultant r so that we can get the final result easily directly from formula.

Complete step-by-step answer:
Electric field: It is defined as the electric property that is associated with each point in space when there is charge present. It is a vector quantity. Its units are(VM)or,(NC)\left (\dfrac{V}{M} \right)\, or, \, \left (\dfrac{N}{C} \right). For positive charge, the electric field lines are radially outwards whereas, for negative charge, the electric field lines are radially inward.
We must be knowing the formula for electric field is –
E=kq(r)r3\overrightarrow {E} =\dfrac {kq\left (\overrightarrow{r} \right)}{{{\left| \overrightarrow{r} \right|}^{3}}} ……………………….. (1)
Where,
E\overrightarrow {E} = electric field due to a point charge,
k=14πε0=9×109k=\dfrac {1} {4\pi {{\varepsilon} _ {0}}} =9\times {{10} ^ {9}}
r\overrightarrow{r} =distance between charge and the given point.
So, from the given data in the question, we can write:
r=3i+4j\overrightarrow{r} =3\mathbf {i} +4\mathbf {j} …………….. (2)
And,
r=32+52=5\left| \overrightarrow{r} \right|=\sqrt{{{3}^{2}}+{{5}^{2}}}=5 ……………. (3)
Putting (2) and (3) in \Rightarrow (1), we get:
E=kq(3i+4j)125\overrightarrow {E} =\dfrac {kq (3\mathbf {i} +4\mathbf {j})} {125}
By putting the values of every variable we are going to get the final solution i.e. the value of electric field:
E=9×109×(2×109)(3i+4j)125\overrightarrow{E}=\dfrac{9\times {{10}^{9}}\times (-2\times {{10}^{-9}})(3\mathbf{i}+4\mathbf{j})}{125}
E=0.424i0.576j\overrightarrow{E}=-0.424\mathbf{i}-0.576\mathbf{j}
E=0.389NC\left| \overrightarrow{E} \right|=0.389\,\dfrac{N}{C}
Hence, we get the value of the electric field at (3m, 4m, 0) as 0.389NC0.389\dfrac {N} {C} .

Note: while dealing with the electrostatics, we must have a clear concept of vector quantity. Because at this point only there are more chances to make mistakes and also while dealing with numerical calculation, put proper data so that calculation mistakes cannot be done. We always refer to solving vector questions through a vector representation graph.