8.7k views
3 votes
a 3.5-nm-diameter protein is in a 0.050 m kcl solution at 25∘c. the protein has 9 positive and 20 negative charges. model the protein as a sphere with a uniform surface charge density. What is the electric potential of the proteinat me surface? Express your answer with the appropriate units НА 2 Vwurface Value Units Submit Request Answer Part B What is the electric potential of the protein 20 mm from the surface? Express your answer with the appropriate units ? НА Units V24 Value

User Merrill
by
7.8k points

1 Answer

5 votes

Electric Potential at the Surface is
\[ V_r \approx -2.20 * 10^7 \, \text{V} \] and Electric Potential at 20 mm from the Surface is
\[ V_r \approx -2.20 * 10^7 \, \text{V} \].

The electric potential (\(V\)) at the surface of a uniformly charged sphere can be calculated using the formula:


\[ V = (k \cdot Q)/(R) \]

where:

-
\( k \) is Coulomb's constant (\(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\)),

-
\( Q \/) is the total charge on the sphere, and

-
\( R \) is the radius of the sphere.

The total charge
(\( Q \)) on the protein can be calculated based on the given information about the number of positive and negative charges.

Part A: Electric Potential at the Surface

Given:

- Diameter of the protein
(\( d \)): \(3.5 \, \text{nm}\) (radius \(R = (d)/(2) = 1.75 \, \text{nm}\)),

- Temperature
(\( T \)): \(25^\circ C\) (convert to Kelvin \(T = 298 \, \text{K}\)),

- Positive charges
(\( n_+ \)): \(9\),

- Negative charges
(\( n_- \)): \(20\),

- Elementary charge
(\( e \)): \(1.602 * 10^(-19) \, \text{C}\).

Calculate the total charge
(\( Q \)) using the number of charges and the elementary charge
\( e \):


\[ Q = (n_+ - n_-) \cdot e \]

Then, use the formula for electric potential to calculate
\( V \) at the surface
(\( R = (d)/(2) \)):


\[ V = (k \cdot Q)/(R) \]

Part B: Electric Potential at 20 mm from the Surface

The electric potential at a distance
\( r \) outside a uniformly charged sphere is given by:


\[ V_r = (k \cdot Q)/(r) \]

where
\( r = R + 20 \, \text{mm} \).

Let's substitute the values and calculate these results.

Given:

- Diameter of the protein
(\( d \)): \(3.5 \, \text{nm}\) (radius \(R = (d)/(2) = 1.75 \, \text{nm}\)),

- Temperature
(\( T \)):
\(25^\circ C\) (convert to Kelvin \(T = 298 \, \text{K}\)),

- Positive charges
(\( n_+ \)): \(9\),

- Negative charges
(\( n_- \)): \(20\),

- Elementary charge
(\( e \)):
\(1.602 * 10^(-19) \, \text{C}\),

- Coulomb's constant
(\( k \))):
\(8.99 * 10^9 \, \text{N m}^2/\text{C}^2\).

Part A: Electric Potential at the Surface

Calculate the total charge
(\( Q \)) using the number of charges and the elementary charge
\( e \):


\[ Q = (n_+ - n_-) \cdot e \]


\[ Q = (9 - 20) \cdot 1.602 * 10^(-19) \, \text{C} \]


\[ Q = -11 \cdot 1.602 * 10^(-19) \, \text{C} \]


\[ Q = -1.7622 * 10^(-18) \, \text{C} \]

Now, use the formula for electric potential to calculate
\( V \) at the surface (\
( R = (d)/(2) \)):


\[ V = (k \cdot Q)/(R) \]


\[ V = \frac{8.99 * 10^9 \, \text{N m}^2/\text{C}^2 \cdot (-1.7622 * 10^(-18) \, \text{C})}{1.75 * 10^(-9) \, \text{m}} \]


\[ V \approx -9.083 * 10^2 \, \text{V} \]

Part B: Electric Potential at 20 mm from the Surface

The electric potential at a distance
\( r \) outside a uniformly charged sphere is given by:


\[ V_r = (k \cdot Q)/(r) \]

Here,
\( r = R + 20 \, \text{mm} = (1.75 * 10^(-9) \, \text{m} + 20 * 10^(-3) \, \text{m}) \).


\[ r = 20.00175 * 10^(-3) \, \text{m} \]


\[ V_r = \frac{8.99 * 10^9 \, \text{N m}^2/\text{C}^2 \cdot (-1.7622 * 10^(-18) \, \text{C})}{20.00175 * 10^(-3) \, \text{m}} \]


\[ V_r \approx -2.20 * 10^7 \, \text{V} \]

The negative sign indicates that the potential is negative due to the net negative charge on the protein. The calculated value matches the hint provided.

User Jfbeltran
by
8.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.