In the figure below, determine the point (other than infinity) at which the elec

Question

In the figure below, determine the point (other than infinity) at which the electric field is zero. (Let q1 = -1.75µC and q2 = 6.70 µC.)

Part 1 of 4 - Conceptualize

Each charged particle produces a field that gets weaker farther away, so the net field due to both charges approaches zero as the distance goes to infinity in any direction. We are asked for the point at which the nonzero fields of the two particles add to zero as oppositely directed vectors of equal magnitude.

Part 2 of 4 - Categorize

The electric field lines are represented by the curved lines in the diagram. The field of positive charge

q2

points radially away from its location. Negative charge

q1

creates a field pointing radially toward its location. These two fields are directed along different lines at any point in the plane except for points along the line joining the particles; the two fields cannot add to zero except at some location along this line. To the right of the positive charge on this line, the fields are in opposite directions but the field from the larger magnitude of the positive charge dominates. In between the two particles, the fields are in the same direction and add together. To the left of the negative charge, the fields are in opposite directions and at some point they will add to zero such that

Part 3 of 4 - Analyze

rLet x represent the distance from the negatively charged particle

q

to the zero-field point to its left. Then

1.00 m + x

is the distance from the positive particle of charge

q+

to this point. At this point, we want to satisfy the condition

so we have

=

.Taking the square root of both sides and cross-multiplying to clear fractions, gives

6.7

1.75

1.96

Part 4 of 4 - Analyze

Solving the equation above for x, we have the distance to the left of the negatively charged particle where

1.05

m.

Explanation / Answer

since both are opposite charged particles hence the field due to these charges in between them never be zero ,only outside the space the field will be zero

let at a distance r to the right of the q2 the field is zero

hence the field due to q1 = Field due to q2

k*q1/r1^2 = k*q2/r2^2

q1/r1^2 = q2/r2^2

1.75/(1+r)^2 = 6.7/r^2

(1+r)^2 / r^2 = 1.75/6.7 = 0.261

(1+r) / r = sqrt(0.261) = + or - 0.51

r = -2.04 m

i.e 1.04 m to the left of the q1 the electirc field is zero

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