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  4. A point charge of 50 μ C is placed in the X Y -plane at a location with radius vector r 0=2 hat i +3 hat j m . The electric field strength and its magnitude at a point with radius vector r =8 hat i -5 hat j m is (ε0=8.85 × 10-12 C 2 / N - m 2)

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Q. A point charge of $50\, \mu C$ is placed in the $X Y$ -plane at a location with radius vector $r _{0}=2 \hat{ i }+3 \hat{ j } m .$ The electric field strength and its magnitude at a point with radius vector $r =8 \hat{ i }-5 \hat{ j } m$ is
$\left(\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} / N - m ^{2}\right)$

TS EAMCET 2018

Solution:

Distance between point charge and observation point is
$x = r - r _{0}=6 \hat{ i }-8 \hat{ j }$
Magnitude of $x =x=\sqrt{6^{2}+8^{2}}$
$x=10 \,m$
$\therefore $ Electric field, $E=\frac{K q}{x^{2}}$
$ =\frac{9 \times 10^{9} \times 50 \times 10^{-6}}{10 \times 10}$
$=4.5 \,kV \, m$