'n Steer Q₁ met 'n lading van -2,5 μC word 1 m vanaf 'n tweede sfeer Q₂ met 'n lading van +6 μC geplaas. Die sferes lê langs 'n regtuig, soos in die diagram hieronder getoon. Punt P is op 'n afstand van 0,3 m links van steer Q₁ geleë, terwyl punt X tussen Q₁ en Q₂ geleë is. Die diagram is nie volgens skaal geteken nie. 8.1 Toon, met behulp van 'n VEKTORDIAGRAM, waarom die netto elektriese veld by punt X nie nul kan wees nie. 8.2 Bereken die netto elektriese veld by punt P as gevolg van die twee geleande sferes Q₁ en Q₂.

NSC Physical Sciences Electrostatics: 'n Steer Q₁ met 'n lading van -2

'n Steer Q₁ met 'n lading van -2,5 μC word 1 m vanaf 'n tweede sfeer Q₂ met 'n lading van +6 μC geplaas - NSC Physical Sciences - Question 8 - 2016 - Paper 1

Question 8

'n Steer Q₁ met 'n lading van -2,5 μC word 1 m vanaf 'n tweede sfeer Q₂ met 'n lading van +6 μC geplaas. Die sferes lê langs 'n regtuig, soos in die diagram hieronde... show full transcript

Worked Solution & Example Answer:'n Steer Q₁ met 'n lading van -2,5 μC word 1 m vanaf 'n tweede sfeer Q₂ met 'n lading van +6 μC geplaas - NSC Physical Sciences - Question 8 - 2016 - Paper 1

Step 1

Toon, met behulp van 'n VEKTORDIAGRAM, waarom die netto elektriese veld by punt X nie nul kan wees nie.

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Answer

In this scenario, we consider the contributions of the electric fields due to both charges Q₁ and Q₂ at point X. Since charge Q₁ has a negative charge of -2.5 μC and charge Q₂ has a positive charge of +6 μC, the electric field created by Q₁ will point towards it, while the electric field due to Q₂ will point away from it.

To illustrate this, we can draw a vector diagram:

The electric field vector due to Q₁, denoted as $E_{Q1}$ , points to the left (towards Q₁).
The electric field vector due to Q₂, denoted as $E_{Q2}$ , points to the right (away from Q₂).

Since both vectors $E_{Q1}$ and $E_{Q2}$ are not equal in magnitude, the resultant electric field at point X cannot be zero, as they act in opposite directions but have different strengths.

Step 2

Bereken die netto elektriese veld by punt P as gevolg van die twee geleande sferes Q₁ en Q₂.

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Answer

To calculate the electric field at point P due to spheres Q₁ and Q₂, we use the formula for the electric field due to a point charge:

$E = k \frac{Q}{r^2}$

where $E$ is the electric field, $k = 9 \times 10^9 \, \text{N·m}^2/ ext{C}^2$ , $Q$ is the charge, and $r$ is the distance from the charge to the point of interest.

For charge Q₁ (-2.5 μC): $E_{Q1} = k \frac{-2.5 \times 10^{-6}}{(0.3)^{2}} = 250 \, \text{N·C}^{-1}\text{ to the left}$
For charge Q₂ (+6 μC): $E_{Q2} = k \frac{6 \times 10^{-6}}{(1-0.3)^{2}} = 31 952.66 \, \text{N·C}^{-1}\text{ to the left}$

To find the total electric field $E_P$ at point P, we sum these two contributions:

$E_P = E_{Q2} + E_{E, -2.5 µC} = 31 952.66 + 250 = 281 952.66 \, \text{N·C}^{-1}\text{ to the left}$

'n Steer Q₁ met 'n lading van -2,5 μC word 1 m vanaf 'n tweede sfeer Q₂ met 'n lading van +6 μC geplaas - NSC Physical Sciences - Question 8 - 2016 - Paper 1

Worked Solution & Example Answer:'n Steer Q₁ met 'n lading van -2,5 μC word 1 m vanaf 'n tweede sfeer Q₂ met 'n lading van +6 μC geplaas - NSC Physical Sciences - Question 8 - 2016 - Paper 1

Toon, met behulp van 'n VEKTORDIAGRAM, waarom die netto elektriese veld by punt X nie nul kan wees nie.

Bereken die netto elektriese veld by punt P as gevolg van die twee geleande sferes Q₁ en Q₂.

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