Answer :
To calculate the force experienced by a positive test charge in an electric field, we use the formula:
[tex]\[ F = qE \][/tex]
where:
- [tex]\( F \)[/tex] is the force experienced by the charge (in newtons),
- [tex]\( q \)[/tex] is the charge (in coulombs),
- [tex]\( E \)[/tex] is the electric field strength (in newtons per coulomb).
Given values:
- [tex]\( q = 8.1 \times 10^{-9} \)[/tex] coulombs,
- [tex]\( E = 9.4 \times 10^7 \)[/tex] newtons/coulomb.
Substitute the given values into the formula:
[tex]\[ F = (8.1 \times 10^{-9}) \times (9.4 \times 10^7) \][/tex]
When we multiply these numbers together:
[tex]\[ F = 0.7614 \text{ newtons} \][/tex]
So, the force experienced by the charge is approximately [tex]\( 0.7614 \)[/tex] newtons.
Now, we need to select the correct answer from the options provided:
A. [tex]\( 1.5 \times 10^{-1} \)[/tex] newtons
B. [tex]\( 9.4 \times 10^{-7} \)[/tex] newtons
C. [tex]\( 7.6 \times 10^{-1} \)[/tex] newtons
D. [tex]\( 8.7 \times 10^{-1} \)[/tex] newtons
The answer closest to our result [tex]\( 0.7614 \)[/tex] is option C:
[tex]\[ \boxed{7.6 \times 10^{-1}} \text{ newtons} \][/tex]
[tex]\[ F = qE \][/tex]
where:
- [tex]\( F \)[/tex] is the force experienced by the charge (in newtons),
- [tex]\( q \)[/tex] is the charge (in coulombs),
- [tex]\( E \)[/tex] is the electric field strength (in newtons per coulomb).
Given values:
- [tex]\( q = 8.1 \times 10^{-9} \)[/tex] coulombs,
- [tex]\( E = 9.4 \times 10^7 \)[/tex] newtons/coulomb.
Substitute the given values into the formula:
[tex]\[ F = (8.1 \times 10^{-9}) \times (9.4 \times 10^7) \][/tex]
When we multiply these numbers together:
[tex]\[ F = 0.7614 \text{ newtons} \][/tex]
So, the force experienced by the charge is approximately [tex]\( 0.7614 \)[/tex] newtons.
Now, we need to select the correct answer from the options provided:
A. [tex]\( 1.5 \times 10^{-1} \)[/tex] newtons
B. [tex]\( 9.4 \times 10^{-7} \)[/tex] newtons
C. [tex]\( 7.6 \times 10^{-1} \)[/tex] newtons
D. [tex]\( 8.7 \times 10^{-1} \)[/tex] newtons
The answer closest to our result [tex]\( 0.7614 \)[/tex] is option C:
[tex]\[ \boxed{7.6 \times 10^{-1}} \text{ newtons} \][/tex]