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The diagram below shows a small sphere X with a charge of +1,5 x 10-16 C and two points A and B at different distances from the sphere X. Where electrons REMOVED or ADDED to sphere X in order to acquire a charge of +1,5 x 10-16 C?

Physics

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Answer

4.5

4.73×10−18 N to the right\text{4.73} \times 10^{-18} \text{ N to the right}

Step-by-step solution

4 steps

Part 4.1

Electrons were removed from sphere X.
A positive charge indicates a deficit of electrons, meaning electrons were taken away from the sphere.

Part 4.2

The electric field at a point is defined as the electrostatic force per unit positive test charge experienced at that point.

Part 4.3

The magnitude of the electric field due to the charged sphere X is greater at point A.
The electric field strength (EE) is inversely proportional to the square of the distance (rr) from the charge (E∝1r2E \propto \frac{1}{r^2}). Since point A is closer to sphere X (10 mm) than point B (30 mm), the electric field is stronger at A.

Part 4.4

Here is a description of the labelled vector diagram:

  • Draw sphere X at the center.
  • Draw point Z to the left of X, 10 mm away.
  • Draw point Y to the right of X, 30 mm away.
  • Sphere X has a positive charge (+1.5×10−16+1.5 \times 10^{-16} C).
  • Point charge Z has a positive charge (+3.2×10−16+3.2 \times 10^{-16} C). Since X and Z are both positive, they repel. The force exerted by Z on X (FXZF_{XZ}) will be directed away from Z, which is to the right.
  • Point charge Y has a negative charge (−2.8×10−16-2.8 \times 10^{-16} C). Since X is positive and Y is negative, they attract. The force exerted by Y on X (FXYF_{XY}) will be directed towards Y, which is to the right.
  • Therefore, both force vectors (FXZF_{XZ} and FXYF_{XY}) originate from X and point to the right.

Part 4.5

4.73×10−18 N to the right\text{4.73} \times 10^{-18} \text{ N to the right}

  1. Identify the given values and convert units.

    Charge of sphere X, QX=+1.5×10−16Q_X = +1.5 \times 10^{-16} C
    Charge of point Z, QZ=+3.2×10−16Q_Z = +3.2 \times 10^{-16} C
    Charge of point Y, QY=−2.8×10−16Q_Y = -2.8 \times 10^{-16} C
    Distance between X and Z, rXZ=10 mm=0.010 mr_{XZ} = 10 \text{ mm} = 0.010 \text{ m}
    Distance between X and Y, rXY=30 mm=0.030 mr_{XY} = 30 \text{ mm} = 0.030 \text{ m}
    Coulomb's constant, k=8.99×109 N⋅m2/C2k = 8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2

  2. Calculate the electrostatic force on X due to Z (FXZF_{XZ}).

    Since both charges are positive, the force is repulsive, directed to the right.

    FXZ=k∣QXQZ∣rXZ2F_{XZ} = \frac{k|Q_X Q_Z|}{r_{XZ}^2} FXZ=(8.99×109 N⋅m2/C2)(1.5×10−16 C)(3.2×10−16 C)(0.010 m)2F_{XZ} = \frac{(8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2)(1.5 \times 10^{-16} \text{ C})(3.2 \times 10^{-16} \text{ C})}{(0.010 \text{ m})^2} FXZ=(8.99×109)(4.8×10−32)1.0×10−4 NF_{XZ} = \frac{(8.99 \times 10^9)(4.8 \times 10^{-32})}{1.0 \times 10^{-4}} \text{ N} FXZ=4.3152×10−18 N (to the right)F_{XZ} = 4.3152 \times 10^{-18} \text{ N (to the right)}

  3. Calculate the electrostatic force on X due to Y (FXYF_{XY}).

    Since X is positive and Y is negative, the force is attractive, directed to the right.

    FXY=k∣QXQY∣rXY2F_{XY} = \frac{k|Q_X Q_Y|}{r_{XY}^2} FXY=(8.99×109 N⋅m2/C2)(1.5×10−16 C)(2.8×10−16 C)(0.030 m)2F_{XY} = \frac{(8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2)(1.5 \times 10^{-16} \text{ C})(2.8 \times 10^{-16} \text{ C})}{(0.030 \text{ m})^2} FXY=(8.99×109)(4.2×10−32)9.0×10−4 NF_{XY} = \frac{(8.99 \times 10^9)(4.2 \times 10^{-32})}{9.0 \times 10^{-4}} \text{ N} FXY=4.1953×10−19 N (to the right)F_{XY} = 4.1953 \times 10^{-19} \text{ N (to the right)}

  4. Calculate the net electrostatic force on sphere X.

    Since both forces are in the same direction (to the right), we add their magnitudes.

    Fnet=FXZ+FXYF_{net} = F_{XZ} + F_{XY} Fnet=(4.3152×10−18 N)+(4.1953×10−19 N)F_{net} = (4.3152 \times 10^{-18} \text{ N}) + (4.1953 \times 10^{-19} \text{ N}) Fnet=(4.3152×10−18 N)+(0.41953×10−18 N)F_{net} = (4.3152 \times 10^{-18} \text{ N}) + (0.41953 \times 10^{-18} \text{ N}) Fnet=(4.3152+0.41953)×10−18 NF_{net} = (4.3152 + 0.41953) \times 10^{-18} \text{ N} Fnet=4.73473×10−18 NF_{net} = 4.73473 \times 10^{-18} \text{ N}

    The net electrostatic force on sphere X is 4.73×10−18 N to the right\boxed{\text{4.73} \times 10^{-18} \text{ N to the right}}.

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