Electrostatics [Electric Field]

[format1998]

Homework Statement

(Please see attachment)
Point P is 5.00 cm from the origin and each charge is 20.0 cm from the origin.

Q₁= -50.0 nC (20.0 cm from the origin on the -x axis)
Q₂= +30.0 nC (20.0 cm from the origin on the +x axis)
Point P is 5.00 cm from the origin (in the +x axis)

Homework Equations

E = kQ/r²
E_NET=E₁+E₂

The Attempt at a Solution

E₁ = (8.99E9 Nm²/C²)((-50E-9 C)/(.25m)²) = -7192 N/C

E₂ = (8.99E9 Nm²/C²)((+30E-9 C)/(.15m)²)= 11986.7 N/C

E_NET=E_NET=E₁+E₂=-7192 N/C + 11986.7 N/C = 4794.67 N/C, -x

According to the answer key, the answer is 19200 N/C, -x. What am I not understanding?

Please help! Any and all help is much appreciated.

 

[berkeman]

Homework Statement

(Please see attachment)
Point P is 5.00 cm from the origin and each charge is 20.0 cm from the origin.

Q₁= -50.0 nC (20.0 cm from the origin on the -x axis)
Q₂= +30.0 nC (20.0 cm from the origin on the +x axis)
Point P is 5.00 cm from the origin (in the +x axis)

Homework Equations

E = kQ/r²
E_NET=E₁+E₂

The Attempt at a Solution

E₁ = (8.99E9 Nm²/C²)((-50E-9 C)/(.25m)²) = -7192 N/C

E₂ = (8.99E9 Nm²/C²)((+30E-9 C)/(.15m)²)= 11986.7 N/C

E_NET=E_NET=E₁+E₂=-7192 N/C + 11986.7 N/C = 4794.67 N/C, -x

According to the answer key, the answer is 19200 N/C, -x. What am I not understanding?

Please help! Any and all help is much appreciated.

To me it looks like the diagram has 20cm for the left charge, but your equations use 25cm. Am I misreading your diagram?

 

[format1998]

I need to find the magnitude of the Electric Field at point P. The charge on the left (-50 nC) is 20 cm from the origin, point P is another 5 cm to the right of the origin. So, the -50 nC charge is 25 cm from point P.