A sphere Q1, with a charge of -2.5 μC, is placed 1 m away from a second sphere Q2, with a charge +6 μC. The spheres lie along a straight line, as shown in the diagram below. Point P is located a distance of 0.3 m to the left of sphere Q1, while point X is located between Q1 and Q2. The diagram is not drawn to scale. 8.1 Show, with the aid of a VECTOR DIAGRAM, why the net electric field at point X cannot be zero. 8.2 Calculate the net electric field at point P, due to the two charged spheres Q1 and Q2.

NSC Physical Sciences Electrostatics: A sphere Q1, with a charge of -2

A sphere Q1, with a charge of -2.5 μC, is placed 1 m away from a second sphere Q2, with a charge +6 μC - NSC Physical Sciences - Question 8 - 2016 - Paper 1

Question 8

A sphere Q1, with a charge of -2.5 μC, is placed 1 m away from a second sphere Q2, with a charge +6 μC. The spheres lie along a straight line, as shown in the diagra... show full transcript

Worked Solution & Example Answer:A sphere Q1, with a charge of -2.5 μC, is placed 1 m away from a second sphere Q2, with a charge +6 μC - NSC Physical Sciences - Question 8 - 2016 - Paper 1

Step 1

8.1 Show, with the aid of a VECTOR DIAGRAM, why the net electric field at point X cannot be zero.

96%

114 rated

Answer

To demonstrate that the net electric field at point X cannot be zero, we need to analyze the direction of the electric fields created by both charges Q1 and Q2.

Since point X is located between the two charges, the electric field due to Q1 ( $E_{Q1}$ ) will be directed towards Q1 because it is negatively charged, while the electric field due to Q2 ( $E_{Q2}$ ) will be directed away from Q2, towards Q1, since it is positively charged.

Thus, both electric field vectors, $E_{Q1}$ and $E_{Q2}$ , are directed towards the left.

This means: $E_{Q1} + E_{Q2} = E_{net}$

As both vectors are in the same direction, the net electric field at point X cannot be zero because you cannot have opposing forces canceling each other out. The fields add up constructively in the same direction.

Step 2

8.2 Calculate the net electric field at point P, due to the two charged spheres Q1 and Q2.

99%

104 rated

Answer

To calculate the net electric field at point P, we first compute the individual electric fields due to Q1 and Q2 at point P.

Electric Field due to Q1

Using Coulomb's law, the electric field ( $E$ ) created by a point charge is given by: $E = k \frac{|Q|}{r^2}$ where:

$k = 9 \times 10^9 \, N \cdot m^2/C^2$ (Coulomb's constant)
$Q = -2.5 \, \mu C = -2.5 \times 10^{-6} \, C$
$r = 0.3 \, m$

Thus, the electric field due to Q1 at point P is: $E_{Q1} = 9 \times 10^9 \frac{|-2.5 \times 10^{-6}|}{(0.3)^2} = 250000 \, N/C$ This field is directed to the left.

Electric Field due to Q2

Similarly, for Q2: $E_{Q2} = k \frac{|Q|}{r^2}$ where:

$Q = +6 \, \mu C = 6 \times 10^{-6} \, C$
$r = 0.7 \, m$ (1 m - 0.3 m)

So, we have: $E_{Q2} = 9 \times 10^9 \frac{|6 \times 10^{-6}|}{(0.7)^2} \approx 31952.63 \, N/C$ This field is also directed to the left.

Net Electric Field at Point P

Now, adding the magnitudes of both fields, we get: $E_{net} = E_{Q1} + E_{Q2} = 250000 \, N/C + 31952.63 \, N/C = 281952.63 \, N/C$ This net field is directed to the left.