1.

A positive charge 50microcoloumb is located in xy plane at a position vector x0=4i+4j. the electric field strength at a point whose position vector is r=10i-4j is

Answer»

We KNOW that,
E = \frac{1}{4 \pi e } \frac{q}{r^{2}}

where r= distance between CHARGE and the point

here, charge is at (4,4)  COORDINATE in xy plane ( taking scalars)
and point is at (10,-4)

Distance between charge and the point = \sqrt{6^{2}+8^{2}} = 10 units

Now,
  E = 9×10^9×50 ×(10^-6)/100
  E = 4.5×10³ V/m
                      



Discussion

No Comment Found

Related InterviewSolutions