IB Physics SL A.2.14 Electric Force Fe

Practise defining electric field strength through the force on a positive test charge and resolving electric force with weight or tension in equilibrium and reference-frame…

Exam points

define electric field direction as the direction of force on a small positive test charge
calculate electric force from equilibrium by resolving tension and weight before assigning its direction
distinguish electric force in a charge's rest frame from additional magnetic effects in another frame

[Maximum number: 3]

A vertical wall carries a uniform positive charge on its surface. This produces a uniform horizontal electric field perpendicular to the wall. A small, positively-charged ball is suspended in equilibrium from the vertical wall by a thread of negligible mass.

Figure for Question A.2.14—Electric force Fe question 1 — IB Physics SL

The thread makes an angle of $30^{\circ}$ with the vertical wall. The ball has a mass of 0.025 kg .

Determine the horizontal force that acts on the ball.

Horizontal force on ball $F=T\sin30^\circ$ .
$T=mg/\cos30^\circ$ .
$F=mg\tan30^\circ=0.025\times9.8\times\tan30^\circ=0.14\ \mathrm{N}$ .

Allow $g=10\ \mathrm{N\ kg^{-1}}$ . Award [3] for a bald correct answer. Award [1 max] for an answer of zero, interpreting the horizontal force as the horizontal component of the net force.

F=mg\tan30^\circ=0.025\times9.8\times\tan30^\circ=0.14\ \mathrm{N}

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A.2.14—Electric force Fe question 1