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# The electric potential at a point (x,y,z) is given by V = a + bx^n volt (a and b are positive constants).  The magnitude of electric field at (1,2,3) is 2 N / C and at (2,3,4) is 8 N / C. The electric field at (3,4,5) is? Pls answer with a brief explanation....

Samyak Jain
333 Points
2 years ago
We know that electric field intensity at a point (x,y,z) is given by E = – [($\dpi{100} \delta$V/$\dpi{100} \delta$x)i + ($\dpi{100} \delta$V/$\dpi{100} \delta$y)j + ($\dpi{100} \delta$V/$\dpi{100} \delta$z)k].
Hence, magnitude of electric field at any point is given by |dV/dr| in r direction.
Here, electric potential varies in x direction only, so we shall substitute dr as dx. Otherwise we need to put dr
as $\dpi{100} \delta$x, $\dpi{100} \delta$y, $\dpi{100} \delta$z separately to get electric field vector.
Now, V = a + b xn  $\dpi{100} \Rightarrow$  dV/dx  =  bn xn – 1.
|dV/dx| at (1,2,3) is bn(1)n – 1  =  2 N/C  $\dpi{100} \Rightarrow$  bn = 2         ...(1)
|dV/dx| at (2,3,4) is bn(2)n – 1  =  8 N/C                          ...(2)
Dividing (2) by (1), we get  (2)n – 1 = 4 = 22
$\dpi{100} \therefore$ n – 1 = 2  or  n = 3. Substitute the value of n in (1) to get b = 2/3.
We get V = a + (2/3) x3
So, |dV/dx| at (x,y,z) is 2 x2.
Thus, magnitude of electric field at (3,4,5) is   2.(3)2  =  18.