The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about 0.075 V exists across the membrane. The thickness of the cell membrane is 7.80 10-9 m. What is the magnitude of the electric field in the membrane?
To find the magnitude of the electric field in the cell membrane, we can use the formula:
Electric field (E) = Potential difference (V) / Distance (d)
Given:
Potential difference (V) = 0.075 V
Distance (d) = 7.80 x 10^-9 m
Plugging in these values into the formula, we get:
E = 0.075 V / 7.80 x 10^-9 m
Now, we need to simplify the units. The volt (V) can be written as kg⋅m²⋅s⁻³⋅A⁻¹, and the meter (m) can be written as kg⋅m. So, we have:
E = (0.075 kg⋅m²⋅s⁻³⋅A⁻¹) / (7.80 x 10^-9 kg⋅m)
Simplifying further, we can cancel out the kg and m units:
E = (0.075 m²⋅s⁻³⋅A⁻¹) / (7.80 x 10^-9)
Now, let's calculate:
E = 0.075 / (7.80 x 10^-9) m²⋅s⁻³⋅A⁻¹
Using scientific notation, we get:
E = 9.615 x 10^6 N/C
So, the magnitude of the electric field in the cell membrane is approximately 9.615 x 10^6 N/C.