Find the electric field strength and the force acting on charge q3 = 2∙10-9 C at the midpoint between two point charges
Find the electric field strength and the force acting on charge q3 = 2∙10-9 C at the midpoint between two point charges q1 = 4∙10-9 C and q2 = -5∙10-9 C, which are located at a distance of 0.6 m from each other. Write the concise statement. Create a diagram showing the vectors of the electric field created by the charges q1 and q2. Using the principle of superposition, the total electric field strength created by charges q1 and q2 can be determined as the sum of vectors E = E1 + E2. Since the electric field vectors in this case have the same direction, it can be written as E = E1 + E2. Calculate the value of the electric field strength.
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+ E2.To find the electric field strength at the midpoint, we need to calculate the individual electric field strengths created by q1 and q2, and then add them together.
The electric field strength, E, created by a point charge, q, at a distance, r, from the charge is given by Coulomb"s law:
\[ E = \frac{k \cdot q}{r^2} \]
Where k is Coulomb"s constant, approximately equal to \(9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\).
Let"s calculate the electric field created by q1 first. Given that q1 = \(4 \times 10^{-9} \, \text{C}\), the distance between q1 and the midpoint is half of the distance between q1 and q2, which is 0.6 m/2 = 0.3 m.
Substituting the values into the formula:
\[ E1 = \frac{(9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \cdot (4 \times 10^{-9} \, \text{C})}{(0.3 \, \text{m})^2} \]
Simplifying:
\[ E1 = \frac{36 \times 10^9}{0.09} \, \text{N/C} \]
\[ E1 = 4 \times 10^{10} \, \text{N/C} \]
Now, let"s calculate the electric field created by q2. Given that q2 = \(-5 \times 10^{-9} \, \text{C}\), the distance between q2 and the midpoint is the same, 0.3 m.
Substituting the values into the formula:
\[ E2 = \frac{(9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \cdot (-5 \times 10^{-9} \, \text{C})}{(0.3 \, \text{m})^2} \]
Simplifying:
\[ E2 = \frac{-45 \times 10^9}{0.09} \, \text{N/C} \]
\[ E2 = -5 \times 10^{10} \, \text{N/C} \]
Now, let"s find the total electric field strength, E, at the midpoint by summing up E1 and E2:
\[ E = E1 + E2 \]
\[ E = (4 \times 10^{10} \, \text{N/C}) + (-5 \times 10^{10} \, \text{N/C}) \]
\[ E = -1 \times 10^{10} \, \text{N/C} \]
Therefore, the electric field strength at the midpoint is \(-1 \times 10^{10} \, \text{N/C}\).
Next, let"s calculate the force acting on charge q3 at the midpoint. The force on a charged particle in an electric field is given by:
\[ F = q \cdot E \]
Where F is the force, q is the charge of the particle, and E is the electric field strength.
Substituting the values into the formula:
\[ F = (2 \times 10^{-9} \, \text{C}) \cdot (-1 \times 10^{10} \, \text{N/C}) \]
Simplifying:
\[ F = -2 \times 10 \, \text{N} \]
Therefore, the force acting on charge q3 at the midpoint is \(-2 \times 10 \, \text{N}\).
To create a diagram showing the vectors of the electric field created by q1 and q2, we can draw two arrows representing the directions and magnitudes of E1 and E2. The arrows should be pointing away from q1 and q2, respectively. The length of the arrows can be proportional to the magnitudes of E1 and E2.