Physics, 01.03.2021 20:50 klu65

Two charges, one +Q and the other −Q, are held a distance d apart. Consider only points on the line passing through both charges and clearly explain your answers to the following: [You can answer this problem without any calculations]. Do not consider any points at infinite distance from the charges. [5 points](a) Find the location of all points, if any, where the electric potential is zero.(b) Find the location of all points, if any, where the electric field is zero.

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Answer from: mchpalmer

a. d/2 mid-way between the charges.

b. d/2 mid-way between the charges.

Explanation:

(a) Find the location of all points, if any, where the electric potential is zero.

Since the charges are of equal magnitude and opposite charge and separated by a distance, d, the electric potential due to the +Q charge is V = kQ/x and that due to the -Q charge is V' = -kQ/(d - x) where x is the point of zero electric potential.

The potential is zero when V + V' = 0, and this can only be midway between the charges. This is shown below

So, kQ/x + [-kQ/(d - x)] = 0

kQ/x - kQ/(d - x) = 0

kQ/x = kQ/(d - x)

1/x = 1/(d - x)

(d - x) = x

d = x + x

d = 2x

x = d/2 which is mid-way between the charges.

(b) Find the location of all points, if any, where the electric field is zero.

Since the charges are of equal magnitude and opposite charge and separated by a distance, d, the electric field due to the +Q charge is E = kQ/x² and that due to the -Q charge is E' = -kQ/(d - x)² where x is the point of zero electric field.

The electric field is zero when E + E' = 0 and this can only be midway between the charges. This is shown below.

So, kQ/x² + [-kQ/(d - x)²] = 0

kQ/x² - kQ/(d - x)² = 0

kQ/x² = kQ/(d - x)²

1/x² = 1/(d - x)²

(d - x)² = x²

d - x = ± x

d = x ± x

d = x - x or x + x

d = 0 or 2x

d = 0 or d = 2x

Since d ≠ 0, d = 2x ⇒ x = d/2 which is midway between the charges.

Answer from: Quest

The geological theory that states that pieces of earth's lithosphere are in constant slow motion is the theory of plate tectonics.

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge