
Uniform electric field of magnitude $100V/m$ in space direct along the line $y = 3 + x$. Find the potential difference between the two points $A(1,3)$ and $B(3,1)$.
A) $100V$
B) $200\sqrt 2 V$
C) $200V$
D) $0$
Answer
137.1k+ views
Hint: Potential difference between two is difference between the value of potential at the given points. Calculate the value of potential at given points in space and find the difference between them. Electric field is basically the gradient of the electric potential.
Complete step by step answer:
Given, $E = 100V/m$ along line $y = 3 + x$.
Slope of given line $y = 3 + x$ is 1.
Means $\tan \theta = 1$ and $\theta = {45^0}$.
Now, we can represent electric field in vector form as in equation below
$\vec E = 100\cos \theta \hat i + 100\sin \theta \hat j$
After putting the value of $\theta $ in the above equation.
$\Rightarrow$ $\vec E = \dfrac{{100}}{{\sqrt 2 }}\hat i + \dfrac{{100}}{{\sqrt 2 }}\hat j$
And points are $A(1,3)$ and $B(3,1)$. Let $\vec r$ be a vector from A to B.
Then, $\Delta \vec r = \overrightarrow {AB} = (\hat i + 3\hat j) - (3\hat i + \hat j)$
$\Rightarrow$ $\Delta \vec r = (1 - 3)\hat i + (3 - 1)\hat j = - 2\hat i + 2\hat j$
We know that relation between potential difference and electric field is given by
$\Rightarrow$ $E = - \dfrac{{\Delta V}}{{\Delta r}}$
Then potential difference is given by $\Delta V = - E.\Delta r$.
So, the potential difference between $A(1,3)$ and $B(3,1)$ is
\[\Delta V = - \left( {\dfrac{{100}}{{\sqrt 2 }}\hat i + \dfrac{{100}}{{\sqrt 2 }}\hat j} \right).\left( { - 2\hat i + 2\hat j} \right)\]
$\Rightarrow$ \[\Delta V = 100\sqrt 2 - 100\sqrt 2 = 0V\]
Hence the correct answer is option D.
Note: If we check then we find that point $A(1,3)$ and $B(3,1)$ lie on line which is perpendicular to line $y = 3 + x$. Then value of potential at points $A$ and $B$ are equal means the potential difference is 0. To find potential difference by electric field magnitude multiply electric field magnitude with the difference between two points along the direction of electric field.
Complete step by step answer:
Given, $E = 100V/m$ along line $y = 3 + x$.
Slope of given line $y = 3 + x$ is 1.
Means $\tan \theta = 1$ and $\theta = {45^0}$.
Now, we can represent electric field in vector form as in equation below
$\vec E = 100\cos \theta \hat i + 100\sin \theta \hat j$
After putting the value of $\theta $ in the above equation.
$\Rightarrow$ $\vec E = \dfrac{{100}}{{\sqrt 2 }}\hat i + \dfrac{{100}}{{\sqrt 2 }}\hat j$
And points are $A(1,3)$ and $B(3,1)$. Let $\vec r$ be a vector from A to B.
Then, $\Delta \vec r = \overrightarrow {AB} = (\hat i + 3\hat j) - (3\hat i + \hat j)$
$\Rightarrow$ $\Delta \vec r = (1 - 3)\hat i + (3 - 1)\hat j = - 2\hat i + 2\hat j$
We know that relation between potential difference and electric field is given by
$\Rightarrow$ $E = - \dfrac{{\Delta V}}{{\Delta r}}$
Then potential difference is given by $\Delta V = - E.\Delta r$.
So, the potential difference between $A(1,3)$ and $B(3,1)$ is
\[\Delta V = - \left( {\dfrac{{100}}{{\sqrt 2 }}\hat i + \dfrac{{100}}{{\sqrt 2 }}\hat j} \right).\left( { - 2\hat i + 2\hat j} \right)\]
$\Rightarrow$ \[\Delta V = 100\sqrt 2 - 100\sqrt 2 = 0V\]
Hence the correct answer is option D.
Note: If we check then we find that point $A(1,3)$ and $B(3,1)$ lie on line which is perpendicular to line $y = 3 + x$. Then value of potential at points $A$ and $B$ are equal means the potential difference is 0. To find potential difference by electric field magnitude multiply electric field magnitude with the difference between two points along the direction of electric field.
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