Constants Part A Two particles having charges of 0.550 nC and 8.80 nC are separa

Question

Constants Part A Two particles having charges of 0.550 nC and 8.80 nC are separated by a distance of 1.90 m At what point along the line connecting the two charges is the net electric field due to the two charges eq to zero? The electric field is zero at a point 0.380 m from 0.550 nC swe Correct Part B Where would the net electric field be zero if one of the charges were negative? Enter your answer as a distance from the charge initially equal 0.550 nC 2.28 Submit Pre ious Answers Request Answer X Incorrect, One attempt remaining; Try Again; no points deducted Provide Feedback Next

Explanation / Answer

a)

The electric field generated by a point charge is

E = (Q/r^2)(1/((4)(pi)(e)))

Where e is the permittivity of free space.

You can set up the two fields and make them equal to each other:

(Q1/r^2)(1/((4)(pi)(e))) = (Q2/(1.9-r)^2)(1/((4)(pi)(e)))

We need to solve for r, where r is the distance from Q1.

the (1/((4)(pi)(e))) cancels on both sides to get:

Q1/r^2 = Q2/(1.9-r)^2 plug in values for Q1 and Q2:

.55/r^2 = 8.8/(1.9-r)^2 (notice the units for charge doesn't matter since they will cancel out as long as they are in the same units) solve for r:

(1.9-r)^2 = 16r^2
=>
1.9-r = 4r
=>
1.9 = 5r
=>
r = 1.9/5 = .38m

so The electric fields are equal when you are .38m away from the .5nC charge and since the electric fields act in opposite directions (both are positively charged particles) the field will equal zero there.

b)

The new position of zero field will be somewhere to the side of one of the charges ( and not between the charges )

Say the 0.55 nC charge is negative

At some distance x from charge the field strength will be + k . 0.55 . 10^–9 / x^2

At the same position the field strength from the other charge will be – k . 8.8 . 10^–9 / ( x + 1.9)^2

These two field strengths must be the same ( to give zero total field ) so

0.55 / x^2 = 8.8 / ( x + 1.9 )^2

0.55* ( x + 1.9 )^2 = 8.8 x^2
or (x+ 1.9)^2 = 16 x^2

x+1.9= 4x

1.9=3x

x=0.633 m

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