Finding electric field at different points

[henchi03]

Homework Statement

A point charge q1 = -6.00nC is at the origin, and a second point charge q2 = +4.00nC is on the x-axis at x = 0.800 m. Find the electric field at each of the following points on the x-axis; a.) 0.200m b.) x = 1.20m c.) x = -0.200 m

Homework Equations

E = kq1q2/r^2

The Attempt at a Solution

i did a little research and saw this:

E = E₁ + E₂

so i did it. E₁ = (9E9)(-6.00E-9)/(0.2)² + E₂ = (9E9)(+4.00E-9)/(0.6)₂

= -1.350x10³ + 1.00X10² = -1.25x10³

but i my book the answer is -1.45X10³

what am i doing wrong?

 

[tiny-tim]

welcome to pf!

hi henchi03! welcome to pf!

E = kq1q2/r^2
…
so i did it. E₁ = (9E9)(-6.00E-9)/(0.2)² + E₂ = (9E9)(+4.00E-9)/(0.6)₂

technically, your E = kq₁q₂/r² equation is a vector equation, not a scalar one

the direction of the vector is an essential part …

it is always outward if q₁q₂ is positive, and inward if q₁q₂ is negative

(the full vector version is E = kq₁q₂r/|r|³)

your r in E₁ is the opposite direction from your r in E₂, so your E₂ should be negative

 

[yinx]

Henchi,

positive charges have E field lines pointing outwards from itself and negative charges have E field lines point inwards.

At point 0.2m, the E field due to Q1 and Q2 are both pointing left.

|E1|= 1350N/C
|E2|= 100N/C

Since both are pointing in the same direction they "reinforce" each other (principle of superposition of E-fields)

therefore you need to sum both E1 and E2 = 1450 and the direction is in the negative x axis.

Hope that helps.

 

[henchi03]

hi henchi03! welcome to pf!


technically, your E = kq₁q₂/r² equation is a vector equation, not a scalar one

the direction of the vector is an essential part …

it is always outward if q₁q₂ is positive, and inward if q₁q₂ is negative

(the full vector version is E = kq₁q₂r/|r|³)

your r in E₁ is the opposite direction from your r in E₂, so your E₂ should be negative

so i should use the vector version of the formula?

r is not the distance between q1 and q2? if it is, how could it be? coz the problem says E is at point 0.2m, it's in between the two charges...

 

[henchi03]

Henchi,

positive charges have E field lines pointing outwards from itself and negative charges have E field lines point inwards.

At point 0.2m, the E field due to Q1 and Q2 are both pointing left.

|E1|= 1350N/C
|E2|= 100N/C

Since both are pointing in the same direction they "reinforce" each other (principle of superposition of E-fields)

therefore you need to sum both E1 and E2 = 1450 and the direction is in the negative x axis.

Hope that helps.

both pointing to the left? how?

0...0.2...0.8

q1...E...q2

isn't q1 pointing to the right?

aren't the charges suppose to be pointing towards E?

 

[tiny-tim]

so i should use the vector version of the formula?

since everything is in a straight line, there's no point …

just remember that the direction (left or right) matters …

draw a little arrow on each of E₁ and E₂, to remind you which way they're pointing, before adding them …

E is really the x-component (E_x) of a vector E

r is not the distance between q1 and q2? if it is, how could it be? coz the problem says E is at point 0.2m, it's in between the two charges...

r is the distance between E and whichever charge we're looking at

q₁ is a positive unit charge at E, and q₂ is one of the other charges

(the positive charge is attracted by one of them, and repelled by the other, so of course both forces are in the same direction)

the standard formula uses q₁ and q₂, but you mustn't confuse that with any charges of the same name!

 

[yinx]

both pointing to the left? how?

0...0.2...0.8

q1...E...q2

isn't q1 pointing to the right?

aren't the charges suppose to be pointing towards E?

I think you are confused, E is not the point at 0.2m.
E means Electric field.

it should look something like this..
let point P be the point at 0.2m

Q1 (-ve) <------ (P) <------- Q2(+ve)

as mentioned, negative charge Q1 will have its own E-field lines pointing towards itself, the arrow left of (P) is the E-field line of Q1 and it is pointing towards Q1 right?

and for positive charge Q2, the field lines will be pointing away from it.. the arrow right of Q2 is the field line of Q2 and it is pointing away from Q2 right?

and if you notice, both arrows are pointed towards left, which means in this context it is pointed towards the -x direction.

therefore the magnitude of E is sum of E1 and E1, and the direction is -x direction (left).

 

[henchi03]

ignore post

 

[henchi03]

I think you are confused, E is not the point at 0.2m.
E means Electric field.

it should look something like this..
let point P be the point at 0.2m

Q1 (-ve) <------ (P) <------- Q2(+ve)

as mentioned, negative charge Q1 will have its own E-field lines pointing towards itself, the arrow left of (P) is the E-field line of Q1 and it is pointing towards Q1 right?

and for positive charge Q2, the field lines will be pointing away from it.. the arrow right of Q2 is the field line of Q2 and it is pointing away from Q2 right?

and if you notice, both arrows are pointed towards left, which means in this context it is pointed towards the -x direction.

therefore the magnitude of E is sum of E1 and E1, and the direction is -x direction (left).

okay last question, for letter b.) the point is at 1.20m

so where would the arrows' direction be?

is (-) charge ALWAYS pointing inwards? pointing to the charge itself? and (+) ALWAYS outwards? so it would be like this?

Q1(-ve)<------------Q2(+ve)------------->(P)

thus E1 = (-) and E2 = (+) ?

 

[yinx]

okay last question, for letter b.) the point is at 1.20m

so where would the arrows' direction be?

is (-) charge ALWAYS pointing inwards? pointing to the charge itself? and (+) ALWAYS outwards? so it would be like this?

Q1(-ve)<------------Q2(+ve)------------->(P)

thus E1 = (-) and E2 = (+) ?

as you can see, in this case, the 2 E fields are "opposing" each other.
Find the magnitude of E1 and E2 and point p, take note which E field have greater magnitude.
minus them off, and the direction is in the direction of the Efield which have greater magnitude.

eg. if the (+ve) has greater mag, the direction is +X direction.