Answer:
(a) 1.6 N/C, due east.
(b) 4 N/C, due east.
Explanation:
The force vector,[tex]\vec {F}[/tex] on the charge, q, in the electric field vector, [tex]\vec{E}[/tex], is
[tex]\vec {F}=\vec{E}q[/tex]
[tex]\Rightarrow \vec{E}=\frac{vec{F}}{q}\;\cdots(i)[/tex]
Let the unit vector towards east is [tex]\hat{i}[/tex], so the unit vector towards west is -
(a) Given that:
[tex]q=+25 \mu C[/tex],
[tex]\vec{F}=40.0 \mu N[/tex], due east
[tex]\Rightarrow \vec{F}=40.0 \mu N (\hat{i})[/tex]
From equation (i)
[tex]\vec{E}=\frac{40.0 \mu N (\hat{i})}{25 \mu C}=1.6(\hat{i}) N/C[/tex]
So, the magnitude of electric field is 1.6 N/C and the direction is towards east.
(b) Given that:
[tex]q=-10.0 \mu C[/tex],
[tex]\vec{F}=40.0 \mu N[/tex], due west
[tex]\Rightarrow \vec{F}=40.0 \mu N (-\hat{i})[/tex]
From equation (i)
[tex]\vec{E}=\frac{40.0 \mu N (-\hat{i})}{-10 \mu C}=-4(\hat{i}) N/C[/tex]
[tex]\Rightarrow \vec{E}=4(\hat{i}) N/C[/tex]
So, the magnitude of electric field is 4 N/C and the direction is towards the east.
