Given that E - -3x-2 - y- hat i - x hat y- kV-m- the work done in moving a - 2 mu C charge from -0- 5- 0- to -2- -1 - 0- by taking the path -y-5-3x12J12mJ12kJ12mu J

Question

Toppr Admin · Accepted Answer

E-3x2-y-j-x-y-kV-mWork done-xA0-x2212-2-x3BC-C-0-5-0- to-xA0-2-x2212-1-0-dw-qEdl-xA0- -xA0- -xA0-xA0-x2212-2-xD7-10-x2212-6-xD7-3x2-y-i-xy-x221A-2-x2212-0-2-x2212-1-x2212-5-2-0-0-2-12-x3BC-J

Given that $E = ((3 x^{2} + y)^i + x^y) k V / m$ , find the work done in moving a $- 2 μ C$ charge from $(0, 5, 0)$ to $(2, - 1, 0)$ by taking the path :
$y = 5 - 3 x$
$12 J$
$12 m J$
$12 k J$
$12 μ J$

$E = (({3 x}^{2} + y)^j + x^y) k V / m$
Work done $= - 2 μ C$
$(0, 5, 0)$ to $(2, - 1, 0)$
$d w = \frac{q E}{d l}$
$=−2×10−6×{(3x2+y)^i+^xy}√(2−0)2+(−1−5)2+(0+0)2=12μJ$

Given that E=((3x2+y)^i+x^y)kV/m, find the work done in moving a −2μC charge from (0,5,0) to (2,−1,0) by taking the path :y=5−3x12J12mJ12kJ12μJ

E=((3x2+y)^j+x^y)kV/mWork done =−2μC(0,5,0) to (2,−1,0)dw=qEdl =−2×10−6×{(3x2+y)^i+^xy}√(2−0)2+(−1−5)2+(0+0)2=12μJ

Given that $E = ((3 x^{2} + y)^i + x^y) k V / m$ , find the work done in moving a $- 2 μ C$ charge from $(0, 5, 0)$ to $(2, - 1, 0)$ by taking the path :
$y = 5 - 3 x$
$12 J$
$12 m J$
$12 k J$
$12 μ J$

$E = (({3 x}^{2} + y)^j + x^y) k V / m$
Work done $= - 2 μ C$
$(0, 5, 0)$ to $(2, - 1, 0)$
$d w = \frac{q E}{d l}$
$=−2×10−6×{(3x2+y)^i+^xy}√(2−0)2+(−1−5)2+(0+0)2=12μJ$