1.

A point charge of 0.009 microcoloumb is placed origin. the intensity of electric field due to this charge at a point (√2,√7,0)

Answer»

A point CHARGE of 0.009μC is PLACED at origin. And we have to find electric field intensity at (√2, √7, 0)

So, position VECTOR , r = (√2 - 0)i + (√7 - 0)j + (0-0)k
r = √2i + √7j + 0k

Now, electric field intensity , E = \frac{KQ}{r^3}\bold{r}
MAGNITUDE of position vector , r = √{√2² + √7² +0²} = 3 unit
Now, E = 9 × 10⁹ × 0.009 × 10⁻⁶ (√2i + √7j + 0k)/(3)³
E = 81/27 (√2i + √7j + 0k)
E = 3(√2i + √7j +0k) N/C

Magnitude of Electric field intensity , E = 9 N/C


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