Question

A positive charge 50 microcoulomb is located in xy plane at position vector R= 4i+4j . The electric field strength E at a point whose position vector is r = 10i-4j is
1) -1.6i-3.6j kV/m
2) 3.6i+1.6j kV/m
3) i-3j kV/m
4) 2.7i-3.6j kV/m

Answer 1

Answer:

As per the data given in the above question.

we need to find the electric field in vector form.

Given,

$charge, Q= 50 μC = 5× {10}^{ - 6} C$

Position vector of point charge

r_[o] =( 4(i) +4(j))

$Position \: vector \: of \: unit \: test \\ charge \: located \\ at \: point P \\ r=( 10\hat[i] - 4\hat(j))</p><p>$

$so \: distance \: between \: point \: an \: charges \:</p><p> r - r_[o] = (10\hat[i] - 4\hat(j)) - (4\hat[i] + 4\hat(j))$

$r\hat-\hat r_(o) = (6\hat[i] - 8\hat(j))$

Now magnitude of

$|r\hat-\hat r_(o) |= |(6\hat[i] - 8\hat(j))| \\ </p><p>=√6²+8²= √36+64= √100= 10$

The electric field strength E at a point

$|E\hat|= \frac{kq}{ {|r\hat-\hat r_(o)|}^{2} } </p><p>=\frac{9 \times {10}^{9} \times 50 \times {10}^{ - 6} }{100}$

$|E\hat|= 4.5 × 10³ \frac{v}{m}$

So ,

$E\hat= 4.5 \times \frac{(6\hat(i) - 8\hat(j)}{10}$

$E\hat=( 0.45 \times(6\hat(i) ) -( 0.45 \times 8\hat(j))$

$E\hat= (2.7\hat(i) - 3.6\hat(j)) \frac{kv}{m}$

Hence , option (d) is correct The electric field strength E at a point

$|E\hat|= (2.7\hat(i) - 3.6\hat(j)) \frac{kv}{m}$