# A particle of charge -2*10^-9C is acted on by a downward electrostatic force of 3*10^-6 N when placed in this field. The gravitational and electrostatic force respectively exerted on a proton placed in this field are???

Sep 16, 2017

$\text{Gravitational force} : F = 1.64 \cdot {10}^{- 26}$ $\text{N}$

$\text{Electrostatic force} : F = 2.4 \cdot {10}^{- 16}$ $\text{N}$

#### Explanation:

Let's use the equation $F = q E$ to find the magnitude of the electric field the particle is in:

$R i g h t a r r o w F = q E$

$R i g h t a r r o w E = \frac{F}{q}$

Considering downwards to be the "negative" direction:

Rightarrow E = frac(- 3 cdot 10^(- 6) " N")(- 2 cdot 10^(- 9) " C")

$\therefore E = 1.5 \cdot {10}^{3}$ ${\text{N" cdot "C}}^{- 1}$

Then, we can apply the same equation to find the electrostatic force that will act on a proton if it were placed in this field:

$R i g h t a r r o w F = 1.6 \cdot {10}^{- 19}$ $\text{C} \cdot 1.5 \cdot {10}^{3}$ ${\text{N" cdot "C}}^{- 1}$

$\therefore F = 2.4 \cdot {10}^{- 16}$ $\text{N}$

Now, the equation for the force due to gravity can be found using the equation $F = m g$:

$R i g h t a r r o w F = 1.673 \cdot {10}^{- 27}$ $\text{kg} \cdot 9.81$ ${\text{m" cdot "s}}^{- 2}$

$\therefore F = 1.64 \cdot {10}^{- 26}$ $\text{N}$