How Do Static Charges Affect X-Ray Procedures?

[treen74]

Hi all,
First time user of this forum so please be gentle with me. I am 4 weeks into a 3 year degree as a radiographer. Radiation physics is a subject that's giving me grief so I've got some questions that I hope you can help me with. I'll crack on.....If you could show me the relevant formulas and workings i will be eternally grateful.

1a) After sliding across an X-ray couch and standing up, a patient is found to have a negative static charge of 1.37nC. Calculate the positive static charge of X-ray couch. Explain your reasoning.

1b) Calculate the force between patient and couch if the are 1m apart.

2a) Xray exposure is made with settings of 85Kvp, 320 mA and 20ms, Calculate the total charge transferred from cathode to anode

2b)Calculate the work done on the electrons

3) 2 Objects charged and placed 1m apart. They experience a force of 10.5N. Calculate Force between charged bodies if they are to be moved 500mm apart.


In previous studies I've thrived by being given an example with workings and then I am able to replicate with future questions. The lecturer in question believes that because its a level 5 course that we should be more investigative so we're only given questions with no answers. So we don't really know when we've tackled questions - if theyre right or wrong !. Anyway - hope you canhelp with any of the above. Apologies if I've got the forum procedure wrong...



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[NateD]

1a] For static charges, the total charge of an isolated system must be conserved. This means that the total charge of the patient and couch together must equal zero. Since the patient has a negative charge of -1.37nC, this means that the couch must have a positive charge of +1.37nC.[1b] The force between the patient and couch can be calculated using Coulomb's law, which states that the force between two charged particles is given by:F = k*(q1*q2)/r^2Where k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the two particles, and r is the distance between them. In this case, q1 = -1.37nC and q2 = +1.37nC, and r = 1m. Plugging these into the equation, we get:F = 8.99 x 10^9 * (-1.37 x 10^-9)*(1.37 x 10^-9) / (1^2)F = 5.45 x 10^-8 N[2a] The total charge transferred from the cathode to the anode can be calculated using the equation:Q = I*tWhere I is the current (320 mA) and t is the exposure time (20 ms). Plugging these into the equation, we get:Q = 320 x 10^-3 * 20 x 10^-3Q = 6.4 x 10^-6 C[2b] The work done on the electrons can be calculated using the equation:W = V*QWhere V is the voltage (85kV) and Q is the total charge (6.4 x 10^-6 C). Plugging these into the equation, we get:W = 85 x 10^3 * 6.4 x 10^-6W = 544 x 10^-3 J[3] The force between the two charged bodies can be calculated using Coulomb's law, which