(C) 8+4 (D) -81 +41 etric field in ...
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(C) 8+4 (D) -81 +41 etric field in a region is given by: E = (4axy Tz)i + (2axIzli+lax? Izk.where The equation ofan equipotential surface will be of the form (B) z=constant / [xy?] (D) None The electric fie constant. The e where a is a positive A) Z=constant / [r327 z=constant / [x4,21 ace

JEE/Engineering Exams
Physics
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Answer At equipotential, surface potential is constant. Therefore, ( E=frac{-d v}{d r}-int d v=int vec{E} cdot d vec{r} ) ( vec{r}=x hat{i}+y hat{j}+z hat{k} ) or ( d vec{r}=d x hat{i}+d y hat{j}+d z hat{k} ) ( int vec{E}(d x hat{i}+d y hat{j}+d z hat{k})=operatorname{cons} tan t ) ( intleft[(4 a x y sqrt{z}) hat{i}+left(2 a x^{2} sqrt{z}right) hat{j}+left(a x^{2} y / sqrt{z}right) hat{k}right] ) ( (d x hat{i}+d y hat{j}+d z hat{k})= )cons tant ( int 4 a x y sqrt{z} d x+int 2 a x^{2} sqrt{z} d y+int frac{a x^{2} y d z}{sqrt{z}}= ) cons tan ( t ) or ( frac{4 a x^{2} y sqrt{z}}{2}+2 a x^{2} y sqrt{z}+2 a x^{2} y sqrt{z}= ) cons tan ( t ) or ( 6 a x^{2} y sqrt{z}= ) cons tan ( t ) or ( sqrt{z}=frac{text { cons } tan t}{6 a x^{2} y} ) or ( z=frac{text { cons tan } t}{x^{4} y^{2}} )
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