The drawing shows a proton with a speed v = 2.9 × 106 m/s moving toward a very long positively-charged plate with a charge density σ = 2.6 × 10-5 C/m2. What is the magnitude of the acceleration of the proton, in meters per squared second, as it approaches the charged plate? **Diagram Explanation**This image depicts an electric field diagram related to electrostatics. It features a vertically aligned row of positive charges represented by plus signs (+) on the right side. These charges are uniformly distributed along a straight strip. To the left of this strip, there is a positive charge symbolized by a circle with a plus sign inside it. An arrow originating from this charge points to the right, indicating the velocity vector (V) of this charge.**Key Elements:**- **Positive Charge (Single):** - Located to the left of the charged strip. - Has a velocity vector, denoted by an arrow labeled "V," pointing towards the strip of charges.- **Charged Strip:** - Positioned vertically and uniformly aligned along the right side. - Consists of positive charges (+) placed at regular intervals along the strip.- **Measurement:** - The distance between the single positive charge and the charged strip is marked as 10 cm, with a horizontal line and double-headed arrow indicating this measurement.This diagram aims to illustrate the interaction of a charged particle with a uniform electric field created by a line of charges.

Given, The speed of the proton, v=2.9×106 m/s The charge density of the plate, σ=2.6×10-5 C/m2 The diagram,Now the electric field due to the positively charged plate is given by, E=σ2ε∘................1 the force is given by, F=qEma=qEa=qEm.................2 on substituting 1 and 2 we get, a=qEm=qσ2mpε∘ where, charge on proton, q=1.6×10-19 C the mass of the proton, mp=1.67×10-27 Kg on putting values we get, a=qσ2mpε∘=1.6×10-19×2.6×10-52×1.67×10-27×8.854×10-12=1.406×1014 m/s2 Hence the acceleration of the proton will be a=1.406×1014 m/s2.