01 Jul, 2021 10:40 PM

## What is the strength of the electric field at the position indicated by the dot in the figure?

E = _____ N/C What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line. &theta; = ____ ⁰&nbsp;

##### Answer
IA
17 Jun, 2021 12:21 PM
distance between dot and upper charge=r1=5\sqrt{2} cm =5\sqrt{2}/100 m E=Electric field due to upper charge=kq/r1^2=1800 N/C(at 45 degree below +x axis) similarly distance between dot and lower charge=r2=5\sqrt{2} cm E=Electric field due to lower charge=kq/r2^2=1800 N/C (at 45 degree above +x axis) Both vertical component of electric field cancel each other because magnitude is equal and direction is opposite and horizontal component add up to give total electric field total field=2*Esin(45)=2545.58 N/C angle=0 degree (along +x axis)