Question

1.
Two point charges (+Q) and (-20) are
fixed on the X-axis at positions a and
2a from origin respectively. At what po-
sitions on the axis, the resultant electric
field is zero
Only x = -2a
both Only x = + 2a
x = sq only
Only a
--
√2a​

Answer 1

Answer:

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Explanation:

Answer 2

Answer:

Let distance of A from (a,0) be x and distance of B from (2a,0) be y.

Point A:

k(+Q)/x² + k(-2Q)/(a+x)² = 0

1/x² - 2/(a+x)² = 0

(a+x)² = 2x²

a ² +x ²+ 2ax =2x²

x²−2ax−a²=0

On simplifying

x=(1± √2)a

x = (1+√2)a

Point B

k(+Q)/( a+y)² + k(-2Q)/(y)² = 0

1/(a+y)² = 2/y²

y²= 2(a+y)²

a ² +y²+ 4ay =y²

y²+4ay+a²=0

y=(−2± √2​)a

y= Not feasible as both are negative.

Thus, as x is found, accordingly the point on x-axis will be (− √2a ,0).

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