Question

Suppose you want to determine the electric field in a certain region of space. You have a small object of known charge and an instrument that measures the magnitude and direction of the force exerted on the object by the electric field. (a) The object has a charge of +25.0 μC and the instrument indicates that the electric force exerted on it is 40.0 μN, due east. What are the magnitude and direction of the electric field? E = (b) What are the magnitude and direction of the electric field if the object has a charge of -10.0 μC and the instrument indicates that the force is 40.0 μN, due west? E = (Chp18,p13GO)

Answer 1

(a) 1.6 N/C, due east.

(b) 4 N/C, due east.

Step-by-step explanation:

The force vector,
`\vec {F}` on the charge, q, in the electric field vector,
`\vec{E}`, is

`\vec {F}=\vec{E}q`

`\Rightarrow \vec{E}=\frac{vec{F}}{q}\;\cdots(i)`

Let the unit vector towards east is
`\hat{i}`, so the unit vector towards west is -

(a) Given that:

`q=+25 \mu C`,

`\vec{F}=40.0 \mu N`, due east

`\Rightarrow \vec{F}=40.0 \mu N (\hat{i})`

From equation (i)

`\vec{E}=\frac{40.0 \mu N (\hat{i})}{25 \mu C}=1.6(\hat{i}) N/C`

So, the magnitude of electric field is 1.6 N/C and the direction is towards east.

(b) Given that:

`q=-10.0 \mu C`,

`\vec{F}=40.0 \mu N`, due west

`\Rightarrow \vec{F}=40.0 \mu N (-\hat{i})`

From equation (i)

`\vec{E}=\frac{40.0 \mu N (-\hat{i})}{-10 \mu C}=-4(\hat{i}) N/C`

`\Rightarrow \vec{E}=4(\hat{i}) N/C`

So, the magnitude of electric field is 4 N/C and the direction is towards the east.