Question

17. Two positive charges of 0.2uc and 0.8uc are placed at a distance of 15 cm apart. At
what point on the line joioning them is the electric field is zero?​

Answer 1

Explanation:

so x=15/√(1/4)+1

x=15/(3/2)

x= 10

here you should note that for like charges put + 1 and for negative put-1 and direction of charge is always from weaker charge

Answer 2

The point on the line joining them where the electric field is zero is 5 cm.

Given: Two positive charges of 0.2 µc and 0.8 µc are placed at a distance of 15 cm apart.

To Find: The point on the line joining them where the electric field is zero.

Solution:

We can solve the numerical using the formula of the electric field which is given by,

            E = kQ / r²                                                            ...(1)

Where E = electric field, k = constant, Q = charge, r = distance.

Coming to the numerical,

Let the distance be 'x' cm.

So, we assume that the electric field will be zero at a distance of x cm from the smaller charge ( 0.2 µC ).

The other distance becomes = ( x - 15 ) cm

From (1), we can say that;

So, the electric field for 0.2 µC is = k × (0.2) / x²

and the electric field for 0.8 µC is = k × (0.8) / (15-x)²

Now, since the sum of the two electric fields is zero, so they are equal.

              k × (0.2) / x²  =  k × (0.8) / (15-x)²

       ⇒    ( 15 - x )²  =  4x²

       ⇒    225 + x² - 30x = 4x²

       ⇒    3x² + 30x - 225 = 0

       ⇒    x²  +  10x - 75 = 0

       ⇒    x²  + 15x - 5x - 75 = 0

       ⇒    x ( x + 15 ) - 5 ( x + 15 ) = 0

       ⇒    x = 5, -15

Since distance cannot be negative so x ≠ -15

Hence, the point on the line joining them where the electric field is zero is 5 cm.

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