Question

wo pairs of charges At which point or points is the electric field zero ? Enter the number or numbers of the points, separated by commas if there are more than one. Enter 0 if there are no points where the field is zero. Figure < 1of1 (a) I 7 8 9 10 Submit 7 89 10 Part B At which point or points is the electric field zero in (Figure 1)(b)? Enter the number or numbers of the points, separated by commas if there are more than one Enter 0 if there are no points where the field is zero.

Answer 1

The electric field is zero at the midpoint between charges of equal magnitude but opposite sign. It will not be zero at any point between charges of different magnitudes. The electric field can be zero at points 8 and 9 in Figure 18.47(a) and Figure 18.48(a). In Figure 18.51(a), there are no points where the electric field is zero. The electric field is also zero at the midpoint between two charges shown in Figure 18.30, Figure 18.33, and Figure 18.22.

Step-by-step explanation:

The electric field at a point due to two charges can be calculated using Coulomb's law. If the electric field is zero, it means that the forces due to the charges cancel each other out. For two charges of the same magnitude but opposite sign, the electric field is zero at the midpoint between them. However, if the charges are not equal, the electric field will not be zero at any point between them. Instead, the electric field will be zero at a point further away from the more negative charge.

In Figure 18.47(a), the electric field is zero at points 8 and 9. In Figure 18.51(a), there are no points where the electric field is zero.

In Figure 18.48(a), the electric field is zero at points 8 and 9. In Figure 18.30, Figure 18.33, and Figure 18.22, the electric field is zero at the midpoint between the two charges.

For the question about charges 91 and 92, if the electric field is zero one-fourth of the way from 91 to q2, it means that the charges 91 and q2 must have the same magnitude but opposite sign.

In the case of an electron orbiting a proton in the hydrogen atom, the angular velocity can be calculated using the formula v = √(kq/r), where k is the electrostatic constant, q is the charge of the electron, and r is the radius of the orbit.

In a system with two parallel conducting plates, the electric field between them can be calculated using the formula E = V/d, where V is the voltage between the plates and d is the distance between them.

Answer 2

Part A. The electric field is zero at point 6

Part B. The electric field is zero at point 9

How to determine the point that the electric field zero?

Consider a scenario with charges at approximate positions
`\( x_1 = 4.5 \)` and
`\( x_2 = 6.5 \)` as estimated from the figure.

The charge on the right side is
`\( q \)` and the left side charge is
`\( 4q \)` because there are four positive signs.

At a neutral point located at a distance
`\( x \)` from point 1:

Electric field due to
`\( 4q \) is \( E_1 = \frac{{k(4q)}}{{(x - x_1)^2}} \)`

Electric field due to
`\( q \)` is
`\( E_2 = \frac{{kq}}{{(x_2 - x)^2}} \)`

At the neutral point, both electric fields are equal in magnitude but opposite in direction:

`\[ E_1 = E_2 \]`

`\[ \frac{{k(4q)}}{{(x - x_1)^2}} = \frac{{kq}}{{(x_2 - x)^2}} \]`

Solving for
`\( x \)`:

`\[ \frac{4}{{(x - 4.5)^2}} = \frac{1}{{(6.5 - x)^2}} \]`

`\[ \frac{{6.5 - x}}{{x - 4.5}} = \sqrt{(1)/(4)} \]`

`\[ \frac{{6.5 - x}}{{x - 4.5}} = (1)/(2) \]`

`\[ 13 - 2x = x - 4.5 \]`

`\[ x = \frac{{13 + 4.5}}{3} = 5.833 \]`

Rounding to the nearest whole number,
`\( x = 5.833 \approx 6 \)`

For Part B, considering the scenario with two charges of opposite signs:

At a neutral point located at a distance
`\( x \)` from point 0:

Electric field due to
`\( 4q \) is \( E_1 = \frac{{k(4q)}}{{(x - x_1)^2}} \)`

Electric field due to
`\( q \) is \( E_2 = \frac{{kq}}{{(x - x_2)^2}} \)`

At the neutral point, both electric fields are equal in magnitude but opposite in direction:

`\[ E_1 = E_2 \]`

`\[ \frac{{k(4q)}}{{(x - x_1)^2}} = \frac{{kq}}{{(x - x_2)^2}} \]`

Solving for
`\( x \)`:

`\[ \frac{4}{{(x - 4.5)^2}} = \frac{1}{{(x - 6.5)^2}} \]`

`\[ \frac{{x - 6.5}}{{x - 4.5}} = \sqrt{(1)/(4)} \]`

`\[ \frac{{x - 6.5}}{{x - 4.5}} = (1)/(2) \]`

`\[ 2x - 13 = x - 4.5 \]`

`\[ x = 8.5 \]`

Rounding to the nearest whole number,
`\( x = 8.5 \approx 9 \)`